1) What is the horizontal component of the ball’s velocity when it leaves Julie\
ID: 1413173 • Letter: 1
Question
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand? 12.7 m/s
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand? 14.12 m/s
3) What is the maximum height the ball goes above the ground? 11.65 m
4) What is the distance between the two girls? 36.6 m
I JUST NEED HELP IN THESE TWO
5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 15 m/s when it reaches a maximum height of 14 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
6) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
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 19 m/s at an angle 48 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.Explanation / Answer
1) Vh (horizontal component) = Vocos() = (19 m/s)cos(48°) = 12.71 m/s.
2) Vv (vertical component) = Vsin() = (19 m/s)sin(48°) = 14.12 m/s.
3)
Use v = v + at. v = 14.12 m/s; a = g = -9.81 m/s²; and v = 0 m/s at the top of the arc. Substitute:
0 = 14.12 m/s - (9.81 m/s²)t, so
(9.81 m/s²)t = 14.12 m/s, so
t = (14.12 m/s)/(9.81 m/s²) = 1.439 sec
Now use this value for t in the general equation for distance with uniform acceleration: x = x + vt + ½at².
Here, x = max height; x = 1.5 m; v = Vv = 14.12 m/s; a = g = -9.81 m/s²; t = 1.439 s. Substitute:
x = 1.5 m + (14.12 m/s)(1.439 s) + 1/2(-9.81 m/s²)(1.439 s)²
x = 11.67 m.
4)
The ball keeps its horizontal speed throughout the time interval. The time to travel is twice the time it takes to get to the top of the arc, because the trajectory is symmetric: t = 2(1.439 s) = 2.878 s.
Use this to find the distance: (12.71 m/s)(2.878 s) = 36.58 m.
5)
If this was phrased properly, Vh = 15 m/s (since Vv = 0, momentarily, when the ball reaches maximum height). This time, use v² = v² + 2ax. We have v = 0 m/s at max height; a = g; and x = 12.5 m. (This is 14 m minus the 1.5 m starting height.)
So v = (v² - 2ax) = (0 - 2(-9.81 m/s²)(12.5 m)) = 15.66 m/s. This is the vertical component of the velocity.
The magnitude of the total velocity is given by (Vh² + Vv²) = (15 m/s)² + (15.66 m/s)² = 21.68 m/s.
6)
Since the girls are 36.58 m apart, use x = vt to find the time of travel
36.58 m = (15 m/s)t, so t = 2.44 s. Now use this in the x = x + vt + ½at² equation. x = 1.5 m;
v = Vv = 15.66 m/s; a = -9.81 m/s²; t = 2.44 s.
Therefore x = 1.5 m + (15.66 m/s)(2.44 s) + 1/2(-9.81 m/s²)(2.43 s)² = 10.54 m.
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