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, [tex]u_y = u \ sin \theta[/tex]
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
[tex]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[/tex]
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;
[tex]t = \frac{X}{V_x} \\\\t = \frac{12}{6} \\\\t = 2 \ s[/tex]
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.
