Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 16 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

Answer 1

Answer:

Explanation:

1.  $V_{x}$ = $V_{0}$ * cos$\alpha$ ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. $V_{y}$ = $V_{0}$ * sin$\alpha$ ⇒ 16* sin32 ≈ 9.4 m/s

3. $y_{max}$ = $\frac{v_{0}^2*sin^2\alpha}{2g}$= $\frac{16^2*sin^232}{2*9.8}$ (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

$y_{max}$ ≈ 3.6677+1.5 ≈ 5.2m

4.  $x_{max}$ = $\frac{v_{0}^2*sin(2\alpha)}{g}$=$\frac{16^2*sin(2*32)}{9.8}$ ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5