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Question -

A player throws a ball upwards with an initial speed of 29.4 m/s.
(a) What is the direction ofacceleration during the upward motion of the ball?
(b) What are the velocity and accelerationof the ball at the highest point of its
motion?
(c) Choose the x = 0 m and t = 0 s tobe the location and time of the ball at its
highest point, vertically downwarddirection 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 riseand after how long does the ball return to the
player’s hands? (Take g = 9.8 m s–2 andneglect air resistance).



Answer -

(a) The accelerationdue 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 zerobut the acceleration due to gravity will be 9.8 m s–2  actingvertically downward.
(c) If we consider the highest point of ball motion as x = 0, t = 0 andvertically 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 pointposition, velocity and acceleration all the three quantities are positive.
(d) Initial speed of the ball, u= -29.4 m/s

Final velocity of theball, v = 0

Acceleration = 9.8 m/s2

Applying in theequation v2 – u2 = 2gs

0 – (-29.4)2 =2 (9.8) s

s = – 864.36/19.6 = –44.1

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

Considering theequation of motion

v = u + at

0 = (-29.4) + 9.8t

t = 29.4/9.8 = 3seconds

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

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