Question

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 75.0 units m (b) number 544.7 units m

Explanation:

Step1: Find initial vertical velocity

The initial vertical velocity $v_{0y}=v_0\sin\theta$, where $v_0 = 48.4$ m/s and $\theta = 28.8^{\circ}$. So $v_{0y}=48.4\times\sin(28.8^{\circ})\approx23.3$ m/s.

Step2: Calculate maximum - height formula

The formula for maximum height of a projectile is $H=\frac{v_{0y}^2}{2g'}$. First, we need to find the acceleration due to gravity on the planet. The range formula on Earth is $R_E=\frac{2v_{0x}v_{0y}}{g_E}$ and on the planet is $R_P=\frac{2v_{0x}v_{0y}}{g_P}$, and we know $R_P = 2.70R_E$, so $g_P=\frac{g_E}{2.70}$. Assuming $g_E = 9.8$ m/s², then $g_P=\frac{9.8}{2.70}\approx3.63$ m/s². Now, substituting $v_{0y}\approx23.3$ m/s and $g_P\approx3.63$ m/s² into the maximum - height formula $H=\frac{v_{0y}^2}{2g_P}=\frac{(23.3)^2}{2\times3.63}\approx75.0$ m.

Step3: Find initial horizontal velocity

The initial horizontal velocity $v_{0x}=v_0\cos\theta$, where $v_0 = 48.4$ m/s and $\theta = 28.8^{\circ}$. So $v_{0x}=48.4\times\cos(28.8^{\circ})\approx42.3$ m/s.

Step4: Calculate time of flight

The time of flight $T=\frac{2v_{0y}}{g_P}$, with $v_{0y}\approx23.3$ m/s and $g_P\approx3.63$ m/s², so $T=\frac{2\times23.3}{3.63}\approx12.8$ s.

Step5: Calculate range

The range $R = v_{0x}T$, substituting $v_{0x}\approx42.3$ m/s and $T\approx12.8$ s, we get $R=42.3\times12.8\approx544.7$ m.

Answer:

(a) 75.0 m
(b) 544.7 m