Problem 4-54 - Projectile - Part 10 - C
A child throws a snowball straight toward a tree with a horizontal velocity of \(18\; m/s\). The tree is \(9.0\; m\) away and the snowball starts from a height of \(1.5\; m\) above the ground.
(a) How long will the snowball take to reach the tree?
(b) At what height above the ground will the snowball hit the tree?
(c) What is the snowball's velocity as it hits the tree?
[Ans. (a) \(0.50\; s\) (b) \(0.3\; m\) (c) \(19\; m/s\; 15^\circ\) down from the horizontal]
Accumulated Solution
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ t = 0.5\; s\) | \(y = v_{0y{^t}} + ½ a_yt^2 \\ y = 1.2\; m \\ h = 0.3\; m \\ v_y = v_{0y} + a_y{^t}\) |
Correct!
Using the result for y that we just worked out we could determine \(v_y\)
\(v_y{^2} + v_{0y}{^2} + 2a_yy = 0^2 + 2(-9.8) (-1.2) = 23.52\)
Therefore \(v_y = 4.8\; m/s\) (we would have got \(4.9\) if we had gone back and used the un-rounded value of \(y\) from the earlier calculation)
Because of the square we have lost the directional information, i.e., \(v_y\) is down.
There is a simple and more direct way and that is to use \(v_y = v_{0y} + a_y{^t}\)