Question

A soccer player takes a free kick from a spot that is 26 m from the goal. The ball leaves his foot at an angle of 35 ∘, and it eventually hits the crossbar of the goal, which is 2.4 m from the ground. At what speed did the ball leave his foot?

Answer 1

Given that after time 't' horizontal displacement of ball is 26 m, while vertical displacement is 2.4 m, while projectile angle is 35 deg, Now using combine equation of projectile motion:

y = x*tan A + (g*x^2)/(2*V0^2*(cos A)^2)

Using given values:

x = 26 m, y = 2.4 m, g = -9.81 m/sec^2 & A = 35 deg

Which gives

2.4 = 26*tan 35 deg - 9.81*26^2/(2*V0^2*(cos 35 deg)^2)

2.4 = 18.205 - 4941.476/V0^2

4941.476/V0^2 = 18.205 - 2.4

V0 = sqrt (4941.476/(18.205 - 2.4))

V0 = 17.68 m/sec = Speed of ball when it leaves his foot

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