A ball is thrown from the ground at an angle of 30 degree. Returns to ground on the same horizontal plane after 3s, what is the velocity of the projection
Answers
Answer:31.76 meters.
Explanation:
Let’s determine the vertical and horizontal components of the initial velocity.
Vertical = 18 * sin 30 = 9 m/s, Horizontal 18 * cos 30
Let’s use the vertical velocity and the initial height in the following equation to determine the time the ball is in the air.
d = vi * t – ½ * g * t^2
d = vertical displacement = Final height – Initial height = 0 – 2 = -2 m
vi = 9, g = 9.8
-2 = 9 * t – 4.9 * t^2
4.9 * t^2 – 9 * t – 2 = 0
t = [9 ± √(81 – 4 * 4.9 * -2)] ÷ 9.8
t = [9 ± √120.2)] ÷ 9.8
t = [9 + √120.2] ÷ 9.8
This is approximately 2.037 seconds.
t = [9 – √120.2 ÷ 9.8
This is approximately = 0.2 seconds.
To determine the horizontal distance the ball moves, multiply its initial horizontal velocity by the first time.
d = 18 * cos 30 * [9 + √120.2)] ÷ 9.8
This is approximately 31.76 meters.