Question

question 1 of 10
you kick a ball with a speed of 14 m/s at an angle of 51°. how far away does the ball land?
a. 1.8 m
b. 19.6 m
c. 21.3 m
d. 3.96 m

Explanation:

Step1: Find the initial horizontal and vertical velocities

The initial speed is $v_0 = 14$ m/s and the angle $\theta=51^{\circ}$. The initial horizontal velocity $v_{0x}=v_0\cos\theta$ and the initial vertical velocity $v_{0y}=v_0\sin\theta$. So $v_{0x}=14\cos(51^{\circ})$ and $v_{0y}=14\sin(51^{\circ})$.

Step2: Find the time of flight

The time - of - flight $t$ of a projectile is given by the equation for vertical motion. When the ball lands, the vertical displacement $y - y_0 = 0$. Using the equation $y - y_0=v_{0y}t-\frac{1}{2}gt^2$, with $y - y_0 = 0$, we get $0 = v_{0y}t-\frac{1}{2}gt^2=t(v_{0y}-\frac{1}{2}gt)$. One solution is $t = 0$ (corresponds to the initial time). The non - zero solution is $t=\frac{2v_{0y}}{g}$, where $g = 9.8$ m/s². Substituting $v_{0y}=14\sin(51^{\circ})$ into the formula, we have $t=\frac{2\times14\sin(51^{\circ})}{9.8}$.

Step3: Find the horizontal range

The horizontal range $R$ is given by $R = v_{0x}t$. Since $v_{0x}=v_0\cos\theta$ and $t=\frac{2v_0\sin\theta}{g}$, then $R=\frac{v_0^{2}\sin(2\theta)}{g}$ (using the double - angle formula $\sin(2\theta)=2\sin\theta\cos\theta$). Substituting $v_0 = 14$ m/s, $\theta = 51^{\circ}$ and $g = 9.8$ m/s², we get $R=\frac{14^{2}\sin(2\times51^{\circ})}{9.8}=\frac{196\times\sin(102^{\circ})}{9.8}$. Since $\sin(102^{\circ})\approx0.9781$, then $R=\frac{196\times0.9781}{9.8}=19.6$ m.

Answer:

B. 19.6 m