Question

1. a player kicks a football from ground level with an initial velocity of 27.0 m/s, 30.0° above the horizontal. find each of the following. assume that forces from the air on the ball are negligible

a. the balls hang time t = 2.76 s
b. the balls maximum height h = 9.30 m
c. the horizontal distance the ball travels before hitting the ground x = 64.6 m

1. the player in the previous problem then kicks the ball with the same speed but at 60.0° from the horizontal. what is the balls hang time, horizontal distance traveled, and maximum height?

Explanation:

Step1: Find vertical - initial velocity

The initial velocity is $v_0 = 27.0$ m/s and the angle $\theta=60.0^{\circ}$. The vertical - initial velocity is $v_{0y}=v_0\sin\theta$. So, $v_{0y}=27.0\sin60^{\circ}=27.0\times\frac{\sqrt{3}}{2}\approx23.4$ m/s.

Step2: Calculate hang - time

The time of flight (hang - time) $T$ for a projectile motion is given by the formula $T = \frac{2v_{0y}}{g}$, where $g = 9.8$ m/s². Substituting $v_{0y}\approx23.4$ m/s, we get $T=\frac{2\times23.4}{9.8}\approx4.78$ s.

Step3: Find maximum height

The formula for the maximum height $H$ of a projectile is $H=\frac{v_{0y}^{2}}{2g}$. Substituting $v_{0y}\approx23.4$ m/s, we have $H=\frac{(23.4)^{2}}{2\times9.8}=\frac{547.56}{19.6}\approx27.9$ m.

Step4: Calculate horizontal distance

The horizontal - initial velocity is $v_{0x}=v_0\cos\theta$. So, $v_{0x}=27.0\cos60^{\circ}=27.0\times\frac{1}{2}=13.5$ m/s. The horizontal distance $x = v_{0x}T$. Substituting $v_{0x}=13.5$ m/s and $T\approx4.78$ s, we get $x=13.5\times4.78 = 64.53$ m.

Answer:

a. The ball's hang - time is approximately $4.78$ s.
b. The ball's maximum height is approximately $27.9$ m.
c. The horizontal distance the ball travels before hitting the ground is approximately $64.53$ m.