Problem 4-57 Projectile - Part 8 - 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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ d = 26.21\; t\) | \(y = -1.5 \;m \\ y = v_{0y{^t}} + ½\; a_yt^2\) |
Correct!
We don't know the final velocity so the only useful equation is: \(y = v_{0y{^t}} + ½\; a_yt^2\)
When the values are substituted into the equation we get a quadratic equation in the variable \(t.\) Do this substitution and pick the equation below that is correct.
(A) \(t^2 – 22.94t - 18.75 = 0\)
(B) \(t^2 + 22.94t - 18.75 = 0\)
(C) \(t^2 – 22.94t + 18.75 = 0\)
(D) \(t^2 + 22.94t + 18.75 = 0\)