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 y ≈ 9.30 m c. the horizontal distance the ball travels before hitting the ground x ≈ 64.6 m 2. 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: Analyze vertical - motion for hang - time

The initial vertical velocity is $v_{0y}=v_0\sin\theta$, where $v_0 = 27.0\ m/s$ and $\theta = 30.0^{\circ}$. In vertical - motion, the time of flight $T$ can be found using the formula $v = v_0+at$. At the end of the motion, the vertical displacement $y - y_0 = 0$. The equation for vertical displacement is $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, and when $y - y_0 = 0$, $0 = v_{0y}t-\frac{1}{2}gt^{2}=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time), and the other is $t=\frac{2v_{0y}}{g}$. Since $v_{0y}=v_0\sin\theta=27.0\sin30.0^{\circ}=27.0\times0.5 = 13.5\ m/s$, then $T=\frac{2v_{0y}}{g}=\frac{2\times13.5}{9.8}\approx2.76\ s$.

Step2: Analyze vertical - motion for maximum height

The formula for the maximum height $H$ in vertical - motion is $v_y^{2}=v_{0y}^{2}-2gH$. At the maximum height, $v_y = 0$. We know $v_{0y}=v_0\sin\theta = 13.5\ m/s$. Rearranging the formula $H=\frac{v_{0y}^{2}}{2g}=\frac{(13.5)^{2}}{2\times9.8}=\frac{182.25}{19.6}\approx9.30\ m$.

Step3: Analyze horizontal - motion for horizontal distance

The initial horizontal velocity is $v_{0x}=v_0\cos\theta$, where $v_0 = 27.0\ m/s$ and $\theta = 30.0^{\circ}$, so $v_{0x}=27.0\cos30.0^{\circ}=27.0\times\frac{\sqrt{3}}{2}\approx23.4\ m/s$. The time of flight is $T = 2.76\ s$. The horizontal distance $x$ is given by the formula $x = v_{0x}T$. So $x=23.4\times2.76\approx64.6\ m$.

For question 2, when the angle is changed to $\theta = 60.0^{\circ}$:

Step4: Calculate new initial vertical and horizontal velocities

$v_{0y}=v_0\sin60.0^{\circ}=27.0\times\frac{\sqrt{3}}{2}\approx23.4\ m/s$ and $v_{0x}=v_0\cos60.0^{\circ}=27.0\times0.5 = 13.5\ m/s$.

Step5: Calculate new hang - time

Using $T=\frac{2v_{0y}}{g}$, $T=\frac{2\times23.4}{9.8}\approx4.78\ s$.

Step6: Calculate new maximum height

Using $H=\frac{v_{0y}^{2}}{2g}$, $H=\frac{(23.4)^{2}}{2\times9.8}=\frac{547.56}{19.6}\approx27.9\ m$.

Step7: Calculate new horizontal distance

Using $x = v_{0x}T$, $x = 13.5\times4.78\approx64.6\ m$.

Answer:

1.
a. $T\approx2.76\ s$
b. $y\approx9.30\ m$
c. $x\approx64.6\ m$
2.
Hang - time: $T\approx4.78\ s$
Horizontal distance: $x\approx64.6\ m$
Maximum height: $y\approx27.9\ m$