Q10. In a basketball match, the referee throws the ball up in the air with an initial speed of 10 m/s.
(a) In what direction is the ball accelerating when it is thrown upwards?
(b) At the highest point of its ascend what is its acceleration and velocity?
(c) Assuming x = 0 and t = 0 to be the location and the time of the basketball at its highest point and the vertically downward direction to be the positive direction of the x axis. What will the signs of velocity, acceleration and position of the basketball be during its downward and upward motion?
(d) To what height does the ball rise? And what is the total air time of the ball?
Answers
Answer:
(d) To what height does the ball rise? And what is the total air time of the ball?
Explanation:
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Solution :-
(a) When the basketball moves upward its acceleration is vertically downwards.
(b) At the highest point of the ball’s ascend the velocity of the ball is 0 and its acceleration ,a = g =9.8 m/s² ( acceleration due to gravity ) in the vertically downward direction.
(c) Taking the above assumption, we get:
(i) During downward motion, x = positive, velocity, v = positive and acceleration, a = g = +ve.
(ii) During upward motion, x = +ve, velocity = -ve and acceleration = g = positive.
(d) Given,
Initial velocity, u = 10 m/s
a = 9.8m/s²
Final velocity, v = 0
Thus, using the third equation of motion, we get:
v²– u² = 2gs
s = (v² – u²) / 2g
s = ( 0 – 10²) / 2 x ( – 9.8 )
s = – 100 / -19.6 = 5.10 m
Therefore the ball attains a maximum height of 5.10m.
Now to find the time of ascent, t
v = u + at
t = (v – u) / a
= -10/-9.8 = 1.02s
Thus, the total time taken by the ball to ascend and come down (air time) = 2 x 1.02 = 2.04 seconds