A stone is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 21.0 m/s, as in the figure. The point of release is h = 43.0 m above the ground.

(a) How long does it take for the stone to hit the ground?
s

(b) Find the stone's speed at impact.
m/s

(c) Find the horizontal range of the stone.
m

Answer:

A stone is thrown from the top of a building upward at an angle of 30.0 degrees to the horizontal?
with an initial speed of 20.0 m/s. If the height of the building is 45.0 m , how long does it take the stown to reach the ground?

20sin 30 = 10m/s
So the stone is moving 10m/s in the y direction. Find how high it goes
At its highest v will = 0

V^2 = Vo^2 + 2ad
0 = (10m/s)^2 + 2(-9.8m/s^2)d
-100m^2/s^2 / -19.6m/s^2 = d
d = 5.102m

We need to find out how long it took to get there
Vf = Vo + at
0 = 10m/s + (-9.8m/s^2)t
(-10m/s) / (-9.8m/s^2) = t
t = 1.02s

So it takes 1.02 s to get up the additional 5.102m. We will add this to the time it takes for free fall

Add the heights together

45m + 5.102m = 50.102m This is how high the stone is off the ground. Initial velocity at this point is 0. Height is -50.102 since it's falling in the negative direction. (down is taken to be negative)

d = Vot + 1/2at^2
-50.102m = 0 + 1/2(-9.8m/s^2)t^2
-50.102m / -4.9m/s^2 = t^2
10.225s^2 = t^2 (take square root)
t = 3.198s

Add the two times together

total time = 3.198s + 1.02s = 4.22s