Question

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.


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Answer 1

Answer:

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, $u_y = u \ sin \theta$

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

$t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-11.328\ \ +/- \ \ \sqrt{(-11.328)^2 - 4(4.9\times -22)} }{2(4.9)}\\\\t = 1.26 \ s$

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;

$t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s$

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.