A ball is projected from point a with a velocity 10m/s perpendicular to the inclined plane
Answers
Let’s determine the vertical and horizontal components of the initial velocity.
Vertical = 10 * sin 60 = 5 * √3 m/s
Horizontal = 10 * cos 60 = 5 m/s
During the time the ball is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. The horizontal velocity is constant. To determine the time, use the following equation.
d = vi * t – ½ * a * t^2, d is the ball’s vertical displacement.
d = final height – initial height
As I look at the figure, I am not sure if the height is 60 meters or 65 meters. I will use 60 meters to show you how to determine the time.
d = 0 – 60 = -60 meters, vi = 5 * √3, a = 9.8
-60 = 5 * √3 * t – ½ * 9.8 * t^2
4.9 * t^2 – 5 * √3 * t – 60 = 0
t = [5 * √3 ± √(75 – 4 * 4.9 * -60)] ÷ 9.8
t = [5 * √3 ± √(1251] ÷ 9.8
t = [5 * √3 + √(1251] ÷ 9.8 = 4.492829837 seconds
t = [5 * √3 – √(1251] ÷ 9.8 = -2.725431054 seconds
The time is 4.492829837 seconds. To determine the range, multiply the time by the initial horizontal velocity.
Range = 5 * 4.492829837 = 22.46414929 meters
This is approximately 22.5 meters. If height is not 60 meters, use the correct height in the quadratic equation. If you use a calculator you will see that sin 60 is the same as √3/2. That is why the initial vertical velocity is 5 * √3 m/s. I hope this is helpful to you.