Problem 4-57 Projectile - Part 4 - A

An astronaut strikes a golf ball on the Moon where the magnitude of the acceleration due to gravity is \(1.6\; m/s^2.\) The ball takes off with a velocity of \(32\; m/s\) at an angle \(35^\circ\) above the horizontal (the moon’s horizontal) and lands in a valley \(15\; m\) below the level where it started. Determine the golf ball's: (a) maximum height (b) time of flight (c) horizontal distance traveled.
[Ans. (a) \(1.1 \times 10^2\; m\)   (b) \(24\; s\)   (c) \(6.2 \times 10^2\; m\)]

Accumulated Solution

Correct!

If \(a_x = 0\) then:

(B) \(x = v_{0x{^t}} + ½ \;a_xt^2 -> x = v_{0x{^t}}\)

(C) \(v_x{^2} = v_{0x}{^2} + 2a_xx -> v_x =v_{0x} \; \text {or} \; v_x \; \text {is constant.}\)

So only \(x = v_{0x{^t}}\) is useful.

The horizontal displacement is:

(A)   \(0\)

(B)   \(1.5\; m\)

(C)   \(d\)