Question

to win the game, a place kicker must kick a football from a point 16 m (17.4976 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 14 m/s at an angle of 55.5° from the horizontal. the acceleration of gravity is 9.8 m/s². by how much vertical distance does the ball clear the crossbar? answer in units of m.

Explanation:

Step1: Find the initial horizontal and vertical velocities

The initial velocity $v_0 = 14$ m/s, and the angle $\theta=55.5^{\circ}$. The initial horizontal velocity $v_{0x}=v_0\cos\theta$ and the initial vertical velocity $v_{0y}=v_0\sin\theta$. So $v_{0x}=14\cos(55.5^{\circ})\approx 7.91$ m/s, $v_{0y}=14\sin(55.5^{\circ})\approx 11.57$ m/s.

Step2: Find the time it takes for the ball to reach the goal

The horizontal - motion is a uniform - motion with $x = v_{0x}t$. Given $x = 16$ m and $v_{0x}\approx7.91$ m/s, we can solve for the time $t=\frac{x}{v_{0x}}=\frac{16}{7.91}\approx2.02$ s.

Step3: Find the vertical position of the ball at time $t$

The vertical - motion is a uniformly - accelerated motion with the equation $y = v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8$ m/s², $v_{0y}\approx11.57$ m/s and $t\approx2.02$ s. So $y=11.57\times2.02-\frac{1}{2}\times9.8\times(2.02)^{2}=11.57\times2.02 - 4.9\times4.0804=23.3714-20.0$ $=3.3714$ m.

Step4: Find the clearance distance

The cross - bar is at $y_{bar}=3.05$ m. The clearance distance $\Delta y=y - y_{bar}=3.3714 - 3.05=0.3214\approx0.32$ m.

Answer:

$0.32$ m