Question

From the top of a building 96ft. high, a ball is thrown directly upward with v=80ft/sec. Find a.) time required to reach the highest point b.) max. height attained. c.) the velocity of the ball when it reaches the ground. :)

Answer 1

initial velociy u= 80ft/sec = 0.3 m/sec.
time taken = time of assent = u/g = 0.3/9.8 = 0.03 sec
maximum ht reached from the tower = u\[u ^{2}/2g \] = \[0.3^{2}/2*9.8\] = .441 m.
Therefore, the total ht of the body from the ground is 96ft + 0.441 = (0.3*96)+0.441 = 24.241m

velocity of the ball when it reaches the ground = \[\sqrt{2gh}\] = \[\sqrt{2*9.8*24.241}\] = \[\sqrt{571.1236}\] = 23.8981 m/s = 76.660ft/sec