A ball is thrown horizontally at 18m/s from the roof of a building lands 28metre from the base of the building. Calculate the height of the building.
To calculate the height of the building, we need to use the equations of motion in the vertical direction. Since the ball is thrown horizontally, its initial vertical velocity is zero. We can use the equation:
s = ut + 1/2at^2
where:
s = displacement in the vertical direction,
u = initial vertical velocity, which is 0,
t = time taken to reach the ground,
a = acceleration due to gravity, which is approximately 9.8 m/s^2.
In this case, the displacement in the vertical direction is the height of the building. The ball is thrown horizontally, so it takes the same time, t, to reach the ground as it would if it were dropped from the same height.
Therefore, we can rearrange the equation:
s = 1/2at^2
Since we want to find the height, s, of the building, we can substitute the given values into the equation:
s = 1/2 * 9.8 * t^2
Now, let's find the value of t. We can use the horizontal distance, 28 meters, and the horizontal velocity, 18 m/s, to calculate the time taken, t. Since the horizontal velocity remains constant throughout the motion, we can use the equation:
s = ut
Rearranging, we have:
t = s/u
t = 28/18
t ≈ 1.56 seconds
Now we can substitute this value of t back into the equation for the height:
s = 1/2 * 9.8 * (1.56)^2
s ≈ 18.106 meters
Therefore, the height of the building is approximately 18.106 meters.
time to fall:
hf=hi-4.9t^2
hf=0
t= sqrt(hi/4.9)
Horizontaldistance:
d=vh*t
28=18*sqrt(hi/4.9)
solvefor hi
hi=4.9*(28/18)^2