A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A player at the bottom of the cliff is 12 m away from the base of the cliff and runs at a maximum speed of 6.0 m/s to catch the ball. Is it possible for the player to catch the ball? Support your answer with calculations.

Please and thankyou!

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff  2 s later. Thus, it is not possible for the player to catch the ball.

Explanation:

Given;

vertical height of the cliff, h = 22 m

velocity of the ball, u = 18 m/s at an angle 39⁰

vertical component of the velocity,

The time for the ball to get to the bottom of the cliff is calculated as;

h = ut + ¹/₂gt²

h = (u sinθ)t + ¹/₂ x 9.8 x t²

22 = (18 sin 39)t + 4.9t²

22 = 11.328t + 4.9t²

4.9t² + 11.328t  - 22 = 0

Solve the above equation with formula method;

a = 4.9, b = 11.328, c = -22

The time for the player to get to the bottom of the cliff;

Given maximum speed, Vx = 6.0 m/s and horizontal distance, X = 12 m;

The ball will get to the bottom of the cliff in 1.26 s, while the player will reach the bottom of the cliff 2 s later. Thus, it is not possible for the player to catch the ball.