A player throws a ball upwards with an initial speed of 29.4 m/s.

**(a)** What is the direction of acceleration during the upward motion of the ball?

**(b)** What are the velocity and acceleration of the ball at the highest point of its motion?

**(c)** Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.

**(d)** To what height does the ball rise and after how long does the ball return to the player’s hands?

(Take g = \( 9.8ms^{-2}\)and neglect air resistance).

Asked by Pragya Singh | 1 year ago | 91

**(a)** The acceleration due to gravity always acts downwards towards the centre of the Earth.

**(b)** At the highest point of its motion the velocity of the ball will be zero but the acceleration due to gravity will be 9.8 m s^{–2}^{ } acting vertically downward.

**(c)** If we consider the highest point of ball motion as x = 0, t = 0 and vertically downward direction to be +ve direction of the x-axis, then

**(i)** During upward motion of the ball before reaching the highest point position ,x = +ve, velocity, v = -ve and acceleration, a = +ve.

**(ii)** During the downward motion of the ball after reaching the highest point position, velocity and acceleration all the three quantities are positive.

**(d)** Initial speed of the ball, u= -29.4 m/s

Final velocity of the ball, v = 0

Acceleration = 9.8 m/s^{2}

Applying in the equation v^{2} – u^{2} = 2gs

0 – (-29.4)^{2} = 2 (9.8) s

s = \( \dfrac{– 864.36}{19.6}\) = – 44.1

Height to which the ball rise = – 44.1 m (negative sign represents upward direction)

Considering the equation of motion

v = u + at

0 = (-29.4) + 9.8t

t =\( \dfrac{ 29.4}{9.8}\) = 3 seconds

Therefore, the total time taken for the ball to return to the player’s hands is 3 +3 = 6s

Answered by Abhisek | 1 year agoThe velocity-time graph of a particle in one-dimensional motion is shown in the figure.

**(a)** \( x (t_2) = x (t_1) + v (t_1) (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a(t_2 – t_1)^2\)

**(b)** \( v(t_2) = v(t_1) + a(t_2 – t_1)\)

**(c)** Vaverage = \( \dfrac{ x(t_2) – x (t_1)}{(t_2 – t_1)}\)

**(d)** aaverage = \( \dfrac{v(t_2) – v (t_1) }{(t_2 – t_1)}\)

**(e)** \( x (t_2) = x (t_1) + v_{av} (t_2 – t_1) +\) (\( \dfrac{1}{2}\)) \( a_{av} (t_2 – t_1)^2\)

**(f)** \( x(t_2) – x (t_1)\) = Area under the v-t curve bounded by t- axis and the dotted lines.

The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

**(a)** t = 0 s to 10 s,

**(b)** t = 2 s to 6 s.

Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of \( 15ms^{-1}\) and \(30ms^{-1}\)Verify that the graph shown in given figure correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = \( 10ms^{-2}\) Give the equations for the linear and curved parts of the plot.

On a long horizontally moving belt figure, a child runs to and fro with a speed 9 km h^{–1} (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^{–1}. For an observer on a stationary platform outside, what is the

**(a)** speed of the child running in the direction of motion of the belt ?.

**(b)** speed of the child running opposite to the direction of motion of the belt?

**(c)** time taken by the child in (a) and (b)?

Which of the answers alter if motion is viewed by one of the parents?

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^{-1}. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s^{-1} and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?