Problem 4-57 Projectile - Part 6 - 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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ d = 26.21\; t\) | - |
Correct!
\(d = 26.21\; t\)
We have 2 unknowns \(d\) and \(t\) so we can go no further with the \(x\)-motion; we must now turn our attention to the \(y\)-motion.
The displacement in the \(y\)-direction is:
(A) \(0\)
(B) \(1.5\; m\)
(C) \(-1.5\; m\)
(D) \(\text{unknown}\)