A stone thrown vertically upwards from the top of a building takes 3s to reach its maximum height . Find
1. The velocity with which it is thrown
2.the maximum height it reaches above the builibg
v = Vi - 9.81 t
0 = Vi - 9.81*3
so Vi = 29.4 m/s
m g h = (1/2) m v^2
so
v^2 = 2(9.81) h
29.4^2 = 19.6 h
h = 44.2 m
Calculate the time taken for stone a to reach it's maximum height
To solve this problem, we can use the basic equations of motion for an object under free fall.
Let's find the answers step by step:
1. The velocity with which the stone is thrown:
When the stone reaches its maximum height, its velocity becomes zero. We can use the equation:
v = u + at
Here, v is the final velocity, u is the initial velocity (which we need to find), a is the acceleration (which is equal to the acceleration due to gravity, approximately -9.8 m/s^2), and t is the time taken (which is 3 seconds in this case).
At the maximum height, v = 0, so the equation becomes:
0 = u + (-9.8 x 3)
Solving for u:
u = 9.8 x 3
u = 29.4 m/s
Therefore, the velocity with which the stone is thrown is 29.4 m/s upwards.
2. The maximum height the stone reaches above the building:
To find the maximum height, we can use the equation:
s = ut + (1/2)at^2
Here, s is the displacement (which is the maximum height above the top of the building), u is the initial velocity (which we found to be 29.4 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken (which is 3 seconds).
Plugging in the values:
s = (29.4 x 3) + (1/2)(-9.8)(3^2)
s = 88.2 + (-44.1)
s = 44.1 m
Therefore, the maximum height the stone reaches above the building is 44.1 meters.