A stone is thrown straight upward from top of building with an initial speed of 35.5m/s, how high does it go from the top of the building? How much time to reach the maximum height? If the building is 45.2 m tall, how much time will it take to hit the ground from when it was initially lunch?
Answers
Given A stone is thrown straight upward from top of building with an initial speed of 35.5m/s, how high does it go from the top of the building? How much time to reach the maximum height? If the building is 45.2 m tall, how much time will it take to hit the ground from when it was initially lunch?
Consider the first case we have
a = 9.8 m /s^2
u = 35.5 m/s
v = 0 m/s
v^2 = u^2 + 2 a s
Since stone is thrown from top of building v = 0
0 = (35.5)^2 + 2 x 9.8 x s
s = 1260.25 / 19.6
s = 64.2 m
Now taking second case we have,
a = - 9.8 m/s^2
u = 35.5 m/s
v = 0
v = u + at
0 = 35.5 + (-9.8)t
t = 35.5 / 9.8
t = 3.62 s
Now for the third case we have
a = - 9.8 m/s^2
u = 0
s = 45.2 + 64.2
s = 109.4 m
t = ?
s = ut + 1/2 at^2
109.4 = 0 + 1/2 x - 9.8 x t^2
- 4.9 t^2 = 109.5
t^2 = - 109.5 / 4.9
t^2 = 22.34
t = 4.72 s
So total time is 3.62 + 4.72 = 8.34 s