Problem 4-54 - Projectile - Part 9 - D
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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0_{x^t}}\) | \(y = v_{0y{^t}} + ½ a_yt^2\) |
| \(t = 0.5\; s\) | \(y = 1.2\; m\) |
| - | \(h = 0.3\; m\) |
Correct!
So now \(y\) can be calculated:
\(y = v_{0y{^t}} + ½ a_yt^2 = 0(0.5) + ½ (-9.8)(0.5)^2 = -1.2\; m\)
Notice the sign is correct as y extends negatively below our chosen origin.
Therefore \(h = 1.5 – 1.2 = 0.3\; m\)
The relation that can be used to find \(v_y\) is:
(A) \(v_y = v_{0y} + a_{y{^t}}\)
(B) \(y = v_{0y{^t}} + ½ \;a_yt^2\)
(C) \(v_y{^2} = v_{0y}{^2} + 2a_yy\)