Question

current attempt in progress
on a distant planet, golf is just as popular as it is on earth. a golfer tees off and drives the ball 2.70 times as far as he would have on earth, given the same initial velocities on both planets. the ball is launched at a speed of 48.4 m/s at an angle of 28.8° above the horizontal. when the ball lands, it is at the same level as the tee. on the distant planet, what are (a) the maximum height and (b) the range of the ball?
(a) number
(b) number

Explanation:

Step1: Find the acceleration due to gravity on the distant planet

The range formula on Earth is $R_{E}=\frac{v_{0}^{2}\sin2\theta}{g_{E}}$, and on the distant planet is $R_{P}=\frac{v_{0}^{2}\sin2\theta}{g_{P}}$. Given that $R_{P} = 2.70R_{E}$, we have $\frac{v_{0}^{2}\sin2\theta}{g_{P}}=2.70\times\frac{v_{0}^{2}\sin2\theta}{g_{E}}$. Canceling out $v_{0}^{2}\sin2\theta$ (since $v_{0}
eq0$ and $\sin2\theta
eq0$), we get $g_{P}=\frac{g_{E}}{2.70}$, where $g_{E} = 9.8\ m/s^{2}$, so $g_{P}=\frac{9.8}{2.70}\ m/s^{2}\approx3.63\ m/s^{2}$.

Step2: Find the maximum - height on the distant planet

The formula for the maximum height of a projectile is $H=\frac{(v_{0}\sin\theta)^{2}}{2g}$. Here, $v_{0} = 48.4\ m/s$, $\theta = 28.8^{\circ}$, and $g = g_{P}\approx3.63\ m/s^{2}$. First, calculate $v_{0}\sin\theta=48.4\times\sin(28.8^{\circ})\approx48.4\times0.482 = 23.33\ m/s$. Then $H=\frac{(23.33)^{2}}{2\times3.63}=\frac{544.29}{7.26}\approx74.97\ m$.

Step3: Find the range on the distant planet

The range formula is $R=\frac{v_{0}^{2}\sin2\theta}{g}$. Calculate $\sin2\theta=\sin(2\times28.8^{\circ})=\sin57.6^{\circ}\approx0.844$. Then $R=\frac{(48.4)^{2}\times0.844}{3.63}=\frac{2342.56\times0.844}{3.63}=\frac{1977.12}{3.63}\approx544.66\ m$.

Answer:

(a) 74.97 m
(b) 544.66 m