Problem 4-57 Projectile - Part 11 - C
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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ d = 26.21\; t\) | \(y = -1.5 \;m \\ y = v_{0y{^t}} + ½\; a_yt^2\\ t^2 - 22.94t - 18.75 = 0 \\ x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\\ t = 23.7 = 24\;s\) |
Correct!
\(t = 11.47 + 12.26 = 23.7 = 24\; s\) (Answer to part b)
The distance traveled can now be found from the \(x\)-motion:
\(d = v_x{^t} = 26.21 × 23.7 = 621 = 6.2×10^2\; m\) (Answer to Part A)
Part (a) asks you to find the maximum height. The condition for maximum height is:
(A) \(v_y = constant\)
(B) \(v_y = 0\)
(C) \(v_x = constant\)
(D) \(v_x = 0\)