Answer:
at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Explanation:
At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest
So here we will say that at the highest point of the path the speed of the ball comes to zero
now by the force diagram we can say that net force on the ball due to gravity is given by

now the acceleration of ball is given as


so at the highest point of the path the acceleration of ball is same as acceleration due to gravity
we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
The correct answer would be C