A ball is thrown straight upward (along the y-axis) with an initial velocity of 16 m/s from the top of a building located 20 m above ground.
Find the maximum height of the ball. In other words, find the ball's maximum distance from the ground.
Be aware that once you find the maximum displacement by using the equations of motion, you will then add 20 m to your answer. This is because the ball was 20 m high at the start.
v = Vi - g t
when v = 0, the ball is at the top.
0 = 16 - 9.8 t
t = 1.633 seconds to the top
y = Yi + Vi t - (1/2)g t^2
y = 20 + 16*1.633 -4.9 * (1.633)^2
y = 20 + 26.13 - 13.07
y = 33.1 m
To find the maximum height of the ball, we need to use the equations of motion. First, let's define the variables:
- Initial velocity (u) = 16 m/s (upward)
- Acceleration (a) = -9.8 m/s^2 (because gravity acts downwards, the acceleration will be negative)
- Displacement (s) = ? (maximum height)
- Final velocity (v) = 0 m/s (at the maximum height, the ball momentarily stops before coming back down)
We can use the following equation of motion to solve for displacement:
v^2 = u^2 + 2as
Since v = 0, the equation becomes:
0 = 16^2 + 2(-9.8)s
Simplifying, we have:
0 = 256 - 19.6s
19.6s = 256
Dividing both sides by 19.6, we find:
s = 256 / 19.6
s ≈ 13.06 m
Therefore, the maximum height of the ball from the ground is approximately 13.06 m.
However, we need to add the initial height of the ball, which is 20 m, to get the actual maximum distance from the ground:
13.06 m + 20 m = 33.06 m
So, the ball's maximum distance from the ground is approximately 33.06 m.