In 1971,Alan Shepard hit two golf-balls on the surface of the Moon (where the acceleration due to gravity is only 1.62 m/s²).If the launch angle was 27 $\circ$ and the ball traveled 400 m,what was the initial speed of the ball? A)28 m/s B)38 m/s C)70 m/s D)801 m/s

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Quizplus · Accepted Answer

The range of a projectile is given by the formula $R = \frac{v^2 \sin 2\theta}{g}$, where $R$ is the range, $v$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. Rearranging for $v$, we get $v = \sqrt{\frac{Rg}{\sin 2\theta}}$. Plugging in the given values ($R = 400$ m, $g = 1.62$ m/s$^2$, and $\theta = 27^\circ$), we find $v \approx 28$ m/s.

In 1971,Alan Shepard Hit Two Golf-Balls on the Surface of the Moon