Question

A girl throws her jackstone ball horizontally out of the window with a velocity of 30m/s. If the window is 3m above the level ground, how far will the ball travel before it hits the ground

Answer 1

Step 1: Write the given

1. Initial velocity is 30 meter per sec
2. Angle is 0° since the jackstone ball is thrown horizontally
3. The window is 3 meters above the ground

Step 2: Determine the formula that will be utilized

1. Since there is x and y component, then the jackstone ball is in projectile motion. 
2. The projectile is assumed ideal since there is no air resistance. 
3. The jackstone ball is thrown horizontally and makes zero degrees with respect to horizontal (theta is equal to 0 degrees).
4. At x component: velocity is constant at any time t
5. At y component: gravity serves as the acceleration like that of free-fall
6. An equation relating x,y, theta, and velocity is: $y=y_o+(tan(\theta))_ox-\frac{g}{2v^2_o(cos(\theta))^2_o}x^2$

Step 3: Find the distance the ball will travel horizontally before it hits the ground

1. change in y or height is equal to -3 (initial height=3 meters and final height is 0 meters which is at the ground)
2. $tan(\theta\times x$ is equal to 0 since tan 0 is equal to 0
3. gravity is equal to 9.81 meter per square second
4. cos(0) is equal to 1
5. The equation will be: $-3=-\frac{9.81}{2 \times 30^2}x^2$
6. $x=\sqrt{\frac{3 \times 2 \times 30^2}{9.81}}$
7. The distance the jackstone ball will travel horizontally is equal to 23.46 meters.

Therefore, the jackstone ball will travel 23.46 meters horizontally before it hits the ground.