a stone is thrown vertically upward an initial velocity of 12m/s from the top of a building . it takes 3.4s for the stone to hit the ground

a. find the height of the building
b. find the maximum height that the stone will reach
c. find the speed of the stone when it reaches the ground

Which question

the free-fall equation is

... h = -4.9 t^2 + 12 t + Hb

a. 0 = -4.9(3.4)^2 + 12(3.4) + Hb

b. Tmax = -12 / (2 * -4.9)
... plug Tmax into the free-fall equation to find Hmax

c. v = g(3.4 - Tmax)

To solve this problem, we can use the equations of motion for vertical motion.

a. Find the height of the building:
We can use the equation: h = ut + (1/2)gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Given:
u = 12 m/s (initial velocity)
t = 3.4 s (time taken to hit the ground)
g = 9.8 m/s^2 (acceleration due to gravity)

Substituting the values into the equation, we get:
h = (12 m/s)(3.4 s) + (1/2)(9.8 m/s^2)(3.4 s)^2

Calculating this expression will give us the height of the building.

b. Find the maximum height that the stone will reach:
To find the maximum height, we can use the equation: v^2 = u^2 + 2gh, where v is the final velocity at the maximum height.
At the maximum height, the final velocity is zero, so we have:
0 = (12 m/s)^2 - 2(9.8 m/s^2)h

Solving this equation will give us the maximum height.

c. Find the speed of the stone when it reaches the ground:
The speed when the stone reaches the ground will be the same as its initial velocity, but in the opposite direction due to gravity. Therefore, the speed is -12 m/s.