Question

To win the game, a place kicker must kick a

Answer 1

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

$v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s$

And the distance to cover is

d = 19 m

So the time taken is

$t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s$

Now we want to find how high the ball is at that time. The initial vertical velocity is

$u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s$

So the vertical position of the ball at time t is

$y(t) = u_y t - \frac{1}{2}gt^2$

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

$y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m$

The crossbar height is 3.05 m, so the difference is

$\Delta h = 3.62 - 3.05 =0.57 m$

So the ball passes 0.57 m above the crossbar.