Physics question

Question

A soccer ball is kicked from midfield and the first bounce hits the goal line 60m away. If the ball is in the air for 3 seconds, what was the speed of the ball when it left the kicker’s boot? Assume the field is level and that wind/air resistance is negligible.

Answer 1

Solution:

You are given time t = 3 sec and the range x = 60 m

the range x = V₀ cos ∝ t , where ∝ is original angle of the ball when it was shot off

then V₀ cos ∝ = x /t = 60 m / 3 s = 20 m/s ---- eq (1)

we also know that the original and final altitude of the ball are equal to zero such that y = y_o = 0

then in the height equation y = -1/2 gt² + V₀ sin ∝ t + y₀ substitute the values

0 = -1/2 gt² + V₀ sin ∝ t + 0 , V₀ sin ∝ t = 1/2 gt² divide by t both sides

V₀ sin ∝ = 1/2 gt = 1/2( 9.8 m/s²)(3 s) = 14.7 m/s ---- eq (2) so you have two equations two unknowns

V₀ cos ∝ = 20 m/s ---- eq (1) , V₀ = 20m/s / cos ∝ substitute this in eq (2)

V₀ sin ∝ = 14.7 m/s ---- eq(2) , ( 20m/s / cos ∝)( sin ∝) = 14.7 m/s , ( 20 m/s)( sin ∝ / cos ∝) = 14.7 m/s

and ( sin ∝ / cos ∝) = tan ∝ = ( 14.7 m/s) / ( 20 m/s) = 0.735

tan^-1tan ∝ = ∝ = tan^-1 ( 0.735) = 36.3º , and V₀ = 20 m/s / cos 36.3º = 24.8 m/s

Check V_o = 14.7 m/s / sin ( 36.3°) = 24.8 m same answer and checks