A diver leaps from the edge of a diving platform into a pool below. The figure above shows the initial position of the diver and her position at a later time. At time $t$ seconds after she leaps, the horizontal distance from the front edge of the platform to the diver's shoulders is given by $x(t)$, and the vertical distance from the water surface to her shoulders is given by $y(t)$, where $x(t)$ and $y(t)$ are measured in meters. Suppose that the diver's shoulders are 11.4 meters above the water when she makes her leap and that $$\frac{dx}{dt} = 0.8 \quad \text{and} \quad \frac{dy}{dt} = 3.6 - 9.8t,$$ for $0 \leq t \leq A$, where $A$ is the time that the diver's shoulders enter the water.
(a) Find the maximum vertical distance from the water surface to the diver's shoulders.
(b) Find $A$, the time that the diver's shoulders enter the water.
(c) Find the total distance traveled by the diver's shoulders from the time she leaps from the platform until the time her shoulders enter the water.
(d) Find the angle $\theta$, $0 < \theta < \frac{\pi}{2}$, between the path of the diver and the water at the instant the diver's shoulders enter the water.

The figure was extracted from an old computer game, called Bang! Bang!
In the game, two competitors control cannons A and B, firing bullets alternately with the objective of hitting the opponent's cannon; to do this, they assign estimated values for the magnitude of the initial velocity of firing ($\left|\overrightarrow{v_0}\right|$) and for the firing angle ($\theta$).
At a certain moment in a match, competitor B must fire; he knows that the bullet fired previously, $\theta = 53^{\circ}$, passed tangentially through point $\boldsymbol{P}$.
In the game, $|\vec{g}|$ equals $10 \mathrm{~m/s}^2$. Consider $\sin 53^{\circ} = 0.8$, $\cos 53^{\circ} = 0.6$ and negligible action of dissipative forces.
Based on the given distances and maintaining the last firing angle, what should be, approximately, the smallest value of $\left|\overrightarrow{v_0}\right|$ that would allow the shot fired by cannon $\mathbf{B}$ to hit cannon $\mathbf{A}$?
(A) $30 \mathrm{~m/s}$.
(B) $35 \mathrm{~m/s}$.
(C) $40 \mathrm{~m/s}$.
(D) $45 \mathrm{~m/s}$.
(E) $50 \mathrm{~m/s}$.

On a day of intense heat, two friends are playing with water from a hose. One of them wants to know how high the water jet reaches from the water outlet when the hose is positioned completely in the vertical direction. The other friend then proposes the following experiment: they position the hose water outlet in the horizontal direction, 1 m in height relative to the ground, and then measure the horizontal distance between the hose and the location where the water hits the ground. The measurement of this distance was 3 m, and from this they calculated the vertical reach of the water jet. Consider the acceleration due to gravity of $10 \mathrm{~m~s}^{-2}$.
The result they obtained was
(A) $1.50 \mathrm{~m}$.
(B) $2.25 \mathrm{~m}$.
(C) $4.00 \mathrm{~m}$.
(D) $4.50 \mathrm{~m}$.
(E) $5.00 \mathrm{~m}$.

The distance between point O and point E is 40 m. As shown in the figure on the right, person A departs from point O and runs along the half-line OS perpendicular to segment OE at a constant speed of 3 m/s, and person B departs from point E 10 seconds after person A starts and runs along the half-line EN perpendicular to segment OE at a constant speed of 4 m/s. The angle formed by the intersection of the segment connecting the positions of persons A and B with segment OE is $\theta$ (in radians). What is the rate of change of $\theta$ at the moment 20 seconds after person A departs? [4 points]
(1) $\frac { 21 } { 290 }$ radians/second
(2) $\frac { 13 } { 290 }$ radians/second
(3) $\frac { 7 } { 290 }$ radians/second
(4) $\frac { 3 } { 290 }$ radians/second
(5) $\frac { 1 } { 290 }$ radians/second

164- In the figure below, a pendulum ball is released from point $A$ and passes through the lowest point of the path with speed $V$. When the ball's speed reaches $\dfrac{\sqrt{2}}{2}\ V$, what angle does the string make with the vertical?
$$\left(g = 10\ \frac{\text{m}}{\text{s}^2},\ \cos 53^\circ = 0.6\right)$$
[Figure: Pendulum of length $1\ \text{m}$ released from point $A$ at $53^\circ$ from vertical]
(Air resistance is neglected.)

• [(1)] $60$ (2) $45$
• [(3)] $37$ (4) $30$

182. Balls A and B are at distance $d$ from each other, and are simultaneously thrown horizontally from the ground surface. Ball A is thrown with speed $30\,\dfrac{\text{m}}{\text{s}}$ in the vertical direction, and ball B is thrown with speed $V_\circ$ at an angle of $45°$ to the horizontal. If both balls collide at a point, how many meters is $d$? (Neglect air resistance, $g = 10\,\dfrac{\text{m}}{\text{s}^2}$)
$$30 \quad (1) \qquad 45 \quad (2) \qquad 60 \quad (3) \qquad 90 \quad (4)$$

Two persons, both of height $h$, are standing at a distance of $h$ from each other. The shadow of one person cast by a vertical lamp-post placed between the two persons is double the length of the shadow of the other. If the sum of the lengths of the shadows is $h$, then the height of the lamp post is
(A) $\frac{\sqrt{3}}{2}h$
(B) $2h$
(C) $\left(\frac{1 + \sqrt{2}}{2}\right)h$
(D) $\left(\frac{\sqrt{3} + 1}{2\sqrt{2}}\right)h$.

Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled 80 km, the triangle became right-angled. When the first ship reaches the port, the second ship was still 120 km from the port. Find the initial distance of the ships from the port.
(A) 240 km
(B) 300 km
(C) 360 km
(D) 180 km

A projectile is thrown from a point O on the ground at an angle $45^\circ$ from the vertical and with a speed $5\sqrt{2}$ m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g = 10$ m/s$^2$.
The value of $x$ is ____.

A ball whose kinetic energy is $E$, is projected at an angle of $45^\circ$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be
(1) $E$
(2) $E / \sqrt{2}$
(3) $E / 2$
(4) zero

A boy playing on the roof of a 10 m high building throws a ball with a speed of $10 \mathrm{~m/s}$ at an angle of $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground?
$$\left[\mathrm{g} = 10 \mathrm{~m/s}^{2}, \sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}\right]$$
(1) 5.20 m
(2) 4.33 m
(3) 2.60 m
(4) 8.66 m

Two stones are projected from the top of a cliff $h$ metres high, with the same speed $u$, so as to hit the ground at the same spot. If one of the stones is projected at an angle $\theta$ to the horizontal then the $\theta$ equals
(1) $u\sqrt{\frac{2}{gh}}$
(2) $\sqrt{\frac{2u}{gh}}$
(3) $2g\sqrt{\frac{u}{h}}$
(4) $2h\sqrt{\frac{u}{g}}$

A projectile can have the same range R for two angles of projection. If $\mathrm{t}_1$ and $\mathrm{t}_2$ be the times of flights in the two cases, then the product of the two time of flights is proportional to
(1) $R^2$
(2) $1/R^2$
(3) $1/R$
(4) R

A particle is projected from a point O with velocity $u$ at an angle of $60^\circ$ with the horizontal. When it is moving in a direction at right angles to its direction at $O$, its velocity then is given by
(1) $\frac{u}{3}$
(2) $\frac{u}{2}$
(3) $\frac{2u}{3}$
(4) $\frac{u}{\sqrt{3}}$

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $v$, the total area around the fountain that gets wet is:
(1) $\pi \frac{v^{4}}{g^{2}}$
(2) $\frac{\pi}{2}\frac{v^{4}}{g^{2}}$
(3) $\pi \frac{v^{2}}{g^{2}}$
(4) $\pi \frac{v^{4}}{g}$

A projectile moving vertically upwards with a velocity of $200 \mathrm{~ms}^{-1}$ breaks into two equal parts at a height of 490 m. One part starts moving vertically upwards with a velocity of $400 \mathrm{~ms}^{-1}$. How much time it will take, after the break up with the other part to hit the ground?
(1) $2\sqrt{10} \mathrm{~s}$
(2) 5 s
(3) 10 s
(4) $\sqrt{10} \mathrm{~s}$

The maximum range of a bullet fired from a toy pistol mounted on a car at rest is $R _ { 0 } = 40 \mathrm {~m}$. What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with uniform velocity $\mathrm { v } = 20 \mathrm {~m} / \mathrm { s }$ on a horizontal surface? $\left( \mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$
(1) $30 ^ { \circ }$
(2) $60 ^ { \circ }$
(3) $75 ^ { \circ }$
(4) $45 ^ { \circ }$

A ball projected from ground at an angle of $45^{\circ}$ just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is:
(1) 4.4 m
(2) 2.4 m
(3) 3.6 m
(4) 1.6 m

A tennis ball (treated as hollow spherical shell) starting from O rolls down a hill. At point A the ball becomes air borne leaving at an angle of $30^{\circ}$ with the horizontal. The ball strikes the ground at B. What is the value of the distance AB? (Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter $= \frac{2}{3}mR^2$)
(1) 1.87 m
(2) 2.08 m
(3) 1.57 m
(4) 1.77 m

The initial speed of a bullet fired from a rifle is $630 \mathrm{~m} / \mathrm{s}$. The rifle is fired at the centre of a target 700 m away at the same level as the target. How far above the centre of the target should the rifle be aimed so that the bullet hits the centre of the target? (Take $g = 10 \mathrm{~m/s}^2$)
(1) 1.0 m
(2) 4.2 m
(3) 6.1 m
(4) 9.8 m

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point $O$ on the ground is $45 ^ { \circ }$. It flies off horizontally straight away from the point $O$. After one second, the elevation of the bird from $O$ is reduced to $30 ^ { \circ }$. Then the speed (in m/s) of the bird is
(1) $20 \sqrt { 2 }$
(2) $20 ( \sqrt { 3 } - 1 )$
(3) $40 ( \sqrt { 2 } - 1 )$
(4) $40 ( \sqrt { 3 } - \sqrt { 2 } )$

An aeroplane flying at a constant speed, parallel to the horizontal ground, $\sqrt { 3 } \mathrm {~km}$ above it is observed at an elevation of $60 ^ { \circ }$ from a point on the ground. If after five seconds, its elevation from the same point is $30 ^ { \circ }$, then the speed (in $\mathrm { km } / \mathrm { hr }$ ) of the aeroplane is
(1) 720
(2) 1500
(3) 750
(4) 1440

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from $30 ^ { \circ }$ to $45 ^ { \circ }$, then the time taken (in $\min$ ) by the car to reach the foot of the tower is
(1) $\frac { 9 } { 2 } ( \sqrt { 3 } + 1 )$
(2) $9 ( \sqrt { 3 } + 1 )$
(3) $18 ( \sqrt { 3 } - 1 )$
(4) $9 ( \sqrt { 3 } - 1 )$

The trajectory of a projectile near the surface of the earth is given as $y = 2x - 9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $g = 10 \text{ m s}^{-2}$:
(1) $\theta_0 = \cos^{-1}\frac{1}{\sqrt{5}}$ and $v_0 = \frac{5}{3} \text{ ms}^{-1}$
(2) $\theta_0 = \cos^{-1}\frac{2}{\sqrt{5}}$ and $v_0 = \frac{3}{5} \text{ ms}^{-1}$
(3) $\theta_0 = \sin^{-1}\frac{1}{\sqrt{5}}$ and $v_0 = \frac{5}{3} \text{ ms}^{-1}$
(4) $\theta_0 = \sin^{-1}\frac{2}{\sqrt{5}}$ and $v_0 = \frac{3}{5} \text{ ms}^{-1}$

Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights, $\mathrm { h } _ { 1 }$ and $\mathrm { h } _ { 2 }$. Which of the following is correct?
(1) $R ^ { 2 } = h _ { 1 } h _ { 2 }$
(2) $R ^ { 2 } = 4 h _ { 1 } h _ { 2 }$
(3) $R ^ { 2 } = 2 h _ { 1 } h _ { 2 }$
(4) $R ^ { 2 } = 16 h _ { 1 } h _ { 2 }$