Problem 4-54 - Projectile - Part 10 - A
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
| \(x\) | \(y\) |
|---|---|
| \(x = v_{0x{^t}} \\ t = 0.5\; s\) | \(y = v_{0y{^t}} + ½ a_yt^2 \\ y = 1.2\; m \\ h = 0.3\; m \\ v_y = v_{0y} + a_y{^t} \\ v_{0y} = 0, \; \text {so}\; v_y = 19\; m/s \\ v = 19\; m/s\; \text {at}\; 15^\circ \; \text{below horiz.}\) |
Correct!
This is the simplest solution (although \(y = v_{0y{^t}} + ½ a_yt^2\) could also be used as well with the value of y we just worked out).
Using this equation, \(v_y= v_{0y} + a_y{^t} = 0 + (-9.8)(0.5) = -4.9\; m/s\)
Now \(v\) can be evaluated:
From Pythagoras
\(v = (v_x{^2} + v_y{^2})^{1/2} = [18^2 + (-4.9)^2]^{1/2} = 19\; m/s\)
\(|\tan \theta| = 4.9/18 = 0.272\;\text{or}\; \theta = 15^\circ\)down from the horizontal.
You have completed this problem.