A ball is thrown upward at 20 metres per second off the top of a building that is 12 m high.
What is the height of the ball above the ground at its highest poin?
v = Vi - g t
0 = 20 - 9.81 t at top when v = 0
t = 2.04 seconds
h = 12 + Vi t -(1/2) g t^2
h = 12 + 20(2.04) -4.9(2.04)^2
h = 32.4 meters
To find the height of the ball above the ground at its highest point, we need to determine how long it takes for the ball to reach its highest point first. We can use the equation of motion to calculate this.
The equation of motion for an object thrown vertically upwards is:
v = u + at
where:
v = final velocity (0 m/s at the highest point, since the ball momentarily stops)
u = initial velocity (20 m/s)
a = acceleration (acceleration due to gravity, which is approximately -9.8 m/s^2)
t = time taken
In this case, we want to find the time taken to reach the highest point, so v = 0 m/s. Rearranging the equation:
0 = 20 m/s - 9.8 m/s^2 * t
Solving for t:
- 20 m/s = - 9.8 m/s^2 * t
t = 20 m/s / 9.8 m/s^2
t ≈ 2.04 seconds
So, it takes approximately 2.04 seconds for the ball to reach its highest point.
Next, we can find the maximum height using the equation:
h = u * t + (1/2) * a * t^2
where:
h = maximum height
u = initial velocity (20 m/s)
t = time taken (2.04 seconds)
a = acceleration (-9.8 m/s^2)
Plugging in the values:
h = 20 m/s * 2.04 s + (1/2) * (-9.8 m/s^2) * (2.04 s)^2
h ≈ 20.4 meters
Therefore, the height of the ball above the ground at its highest point is approximately 20.4 meters.