Football physics pearson question

Question

A football is kicked from a point on the ground that is a horizontal distance d from the goalpost. For the attempt to be successful, the ball must clear the crossbar, 10 ft (about 3.05 m) above the ground. The ball leaves the kicker's foot with an initial speed of 20.0 m/s at an angle of 30.0∘ above the horizontal. Calculate the speed of travel of the football when it reaches the top of the goal.

Answer 1

Standard problem, given V_o and θ, find V_x and V_y (V_x = V_o cosθ = 20.0 m/s (cos 30.0°) =17.32 m/s, V_y =

V_o sinθ = 20.0 m/s (sin 30.0°) = 6.34 m/s) . Find the final V_y at h = 3.05 m by the use of the vertical equation

V² - V_o² = 2gh , this would be coming down from the maximum height where V_o would be zero so

V² - 0 = 2(-9.8 m/s² )( 3.05 m ) = 40,22 m²/s² or V_{final at the goal post height} = - 6.34 m/s.

The speed would be the result of V_x ( which doesn't change for the entire flight ) and V_{final at the goal} or the result of the pythagorean theorem V² = (17.32 m/s )² + ( -6.34 m/s )², V = 18.44 m/s

Answer 2

If it just has to clear that height then that means it gets to that height, the vertical velocity is zero, then it starts accelerating back down.

So if we look at the ball in that moment right before it falls back down and the vertical velocity is zero, then the only velocity is the horizontal velocity. That is a constant because there are no forces (F=m*a so no forces means no acceleration in that direction which means no change in velocity in that direction)

Since velocity is constant along the horizontal and that's the only component of velocity at the top of the ball's travel, the Velocity = V_horizontal at that moment and since V_horizontal is constant it's V*cos(30) = 20*cos(30) which = 17.32 m/s.

However, if you want to know just when it clears the bar (so the ball goes higher than the goal rather than just making it over, which is the case here because the ball should go around 5.1 m high according to the distance formula) you must include a vertical component as well. in this case V_{f, vertical}² = V_{i, vertical}² + 2*a*d which you plug in d = 3.05m, a = -9.8m/s/s, and V_{i, vertical} is 20*sin(30) which is 10 m/s.

Solve that and get V_{f, vertical} is 6.3419 m/s. To get total velocity you take sqrt(V_v²+V_h²) = sqrt(6.34²+17.32²) and get around 18.44 m/s total.