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

##### Solution :-

(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/s2

Applying in the equation v2 – u2 = 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 ago

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