Question

A 58.7-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high.

A) How far does the ball fall in the first 3.08 seconds of flight? = 46.5 m

B) What is the speed of the ball after it has traveled 2.10 m downward? = 6.42 m/s

C) What is the speed of the ball 3.08 s after it is released? = 30.2 m/s

D) If the ball is thrown vertically upward from the top of the tower with an initial speed of 4.85 m/s, where will it be after 2.31 s? Give a positive answer for a location above the top of the tower, or a negative answer for a location below the top of the tower.

I only need help with the last part D. I provided the answers to the others just in case they are needed.

Answer 1

starts at 55m but call it zero because relative to top of tower is asked
v = 4.85 - 9.81 t
h = 4.85 t - 4.9 t^2

= 4.85*2.31 - 4.9(2.31^2)
= 11.2 - 26.1
= -14.9
about 15 meters below top of tower

Answer 2

A. d = 0.5g*t^2.

B. V^2 = Vo^2 + 2g*d. Vo = Initial velocity.
V^2 = 0 + 19.6*2.10 = 41.16,
V = 6.42 m/s.

C. V = Vo + g*t. = 0 + 9.8*3.08 =

D. V = Vo + g*Tr.
0 = 4.85 - 9.8Tr,
Tr = 0.50 s. = Rise time.
Tf = Tr = 0.50 s. = Fall time from max ht. to top of tower.

Tr+Tf = 0.50 + 0.50 = 1.0 s = Time at which ball returned to top of tower.

h = ho - (Vo*t + 0.5g*t^2). t = 2.31-1 = 1.31s.

h = 55-(4.85*1.31+4.9*(1.31)^2 = 40.2m
Above gnd.

40.2 - 55 = -14.8 m = 14.8 m below top of tower.