Problem 4-57 Projectile - Part 13 - B
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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ d = 26.21\; t\) | \(y = -1.5 \;m \\ y = v_{0y{^t}} + ½\; a_yt^2\\ t^2 - 22.94t - 18.75 = 0 \\ x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\\ t = 23.7 = 24\;s\\ = 24\;s \\ \text {For max height} \; v_y= 0 \\ \text{Then}\; y_{max} = 105 = 1.1 \times 10^2\; m\) |
Correct!
(C) \(v_y{^2} = v_{0y{^2}} + 2a_yy\)
\(0^2 = (18.35)^2 + 2(-1.6)y_{max}\)
Therefore \(y_{max} = 105 = 1.1\times 10^2\; m\) (Answer to part a)
You have completed this problem.