You stand at the top of a tall building with a stopwatch. You drop a rock off the side of the building, and it takes the rock 3.2 seconds to hit the ground. (Assume no air resistance.) solved for: [please help check!] the building is 50.176 m tall the impact speed is 31.36 m/s If you were standing at the bottom of this building, at what speed would you have to throw the rock for it to reach the top of the building?

50.176m ; 31.36m/s ; 31.36m/s

Explanation:

Given the following :

Time (t) = 3.2s

Height(s) of the building

The initial velocity (u) will be 0

Using the equation:

S = ut + 0.5at^2

Acceleration due to gravity (a) = g = +9.8m/s^2 (downward)

S = 0*t + 0.5(9.8)(3.2^2)

S = 0 + 50.176

S = 50.176m

B.) speed impact when it touches the ground:

v^2 = u^2 + 2aS

where v is the final Velocity

v^2 = 0^2 + 2(9.8)(50.176)

v^2 = 0 + 983.4496

v = √983.4496

v = 31.36m/s

C) Speed of throw from the ground to reach the top of the building

Here we need the initial velocity

Height (s) = 50.176m

Acceleration due to gravity g = a = - 9.8m/s ( upward)

Using the third equation of motion:

S = ut + 0.5at^2

50.176 = u*3.2 + 0.5(-9.8)(3.2^2)

50.176 = 3.2u - 50.176

50.176 + 50.176 = 3.2u

100.352 = 3.2u

u = 100.352 / 3.2

u = 31.36m/s

answer: a. same unit

explanation: