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.

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

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.

To solve part D, we need to calculate the height of the ball after 2.31 seconds when it is thrown vertically upward from the top of the tower with an initial speed of 4.85 m/s.

The equation for the height of an object in free fall is given by h = h0 + v0t + (1/2)at^2, where:
- h is the final height
- h0 is the initial height
- v0 is the initial velocity
- t is the time
- a is the acceleration due to gravity (which is approximately -9.8 m/s^2 on Earth)

In this case:
- h0 is 55.0 m (height of the tower)
- v0 is 4.85 m/s (initial velocity)
- t is 2.31 s (time)
- a is -9.8 m/s^2 (acceleration due to gravity)

Plugging these values into the equation, we get:
h = 55.0 + (4.85)(2.31) + (1/2)(-9.8)(2.31)^2

Calculating this expression, we find:
h ≈ 55.0 + 11.2135 - 25.7371

Simplifying the expression, we get:
h ≈ 40.4764

So, the ball will be approximately 40.4764 m above the top of the tower after 2.31 s.

To answer part D, we need to calculate the height of the ball after 2.31 seconds when it is thrown vertically upward from the top of the tower with an initial speed of 4.85 m/s.

We can use the kinematic equation for vertical motion:

h = h0 + v0t - (1/2)gt^2

where:
h is the final height
h0 is the initial height (in this case, the height of the tower)
v0 is the initial velocity (4.85 m/s)
t is the time (2.31 seconds)
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Let's plug in the values into the equation:

h = 55.0 m + (4.85 m/s)(2.31 s) - (1/2)(9.8 m/s^2)(2.31 s)^2

Simplifying the equation:

h = 55.0 m + 11.2035 m - 6.76779 m

h = 59.43571 m

Therefore, the ball will be 59.43571 meters above the top of the tower after 2.31 seconds.