7, A all is thrown from the top of a building with inital velocity of 25.9sec st

Question

7, A all is thrown from the top of a building with inital velocity of 25.9sec staright upwards, at an inital height of 100 meters above the ground. Determine the following:

7.1, the time needed for the ball to reach its maximum height

7.2, the maximum height

7.3, the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant

7.4, the time needed for the ball to reach the ground

7.5, the velocity and position of the ball at t=8.00 seconds

7.6, ***suppose after the ball reaches the ground ,it bounces back straight upwards at the speed of half of the speed that it hits the ground, calculate the maximum height that the ball bounces.

Explanation / Answer

A. Vf=Vi + gt
0= 25.9 + 9.8t
t= 2.64 sec

B. PE=KE
.5mv^2=mgh <--- masses cancel
.5(25.9^2) = 9.8h
h=34.225 m

Total h=34.255 +100 =14.255 m

C Vf= Vi + gt
-25.9 = 25.9- 9.8t
t=5.285 s
Velocity 25.9 m/s

becoz PE + KE initial =PE + KE after

D. Vf^2 = Vi^2 + 2ad
Vf^2 = -25.9 ^2 + 2(-9.8)(1000)
Vf=-137.58 m/s

Vf= Vi + at
-137.58 = -25.9 -9.8t
t=11.39 sec
Total time = 11.39+ 2.64 =14.3 sec
E. Vf=Vi + at
Vf= =25.9 -9.8(11.39-8-5.285)=44.71 m/s

Vf^2=Vi^2 + 2ad
Vf=44.71
Vi=25.9
d=67.26 m

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