Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc

Question

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.0 meters above the ground with an initial speed of 10 m/s at an angle 30 degrees with respect to the horizontal. Sarah catches the ball 1.0 meters above the ground. (a) What is the horizontal component of the balls velocity right before Sarah catches it? (b) What is the vertical component of the balls velocity right before Sarah catches it? (c) What is the time the ball is in the air? (d) What is the distance between the two girls? (e) After catching the balk arah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position Assume the ball leaves Sarah's hand a distance 1.0 meters above the ground, reaches a maximum height of 6 m above the ground, and takes 1.115 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand? (f How high above the ground will the ball be when it gets to Julie?

Explanation / Answer

1) Vh (horizontal component) = Vcos() = (10 m/s)cos(30°) = 8.66 m/s.

2) Vv (vertical component) = Vsin() = (10 m/s)sin(30°) = 5 m/s.

3) Use v = v + at. v = 5 m/s; a = g = -9.81 m/s²;

and v = 0 m/s at the top of the arc. Substitute:
0 = 5 m/s - (9.81 m/s²)t, so
(9.81 m/s²)t = 5 m/s, so
t = .5097 s (to 3 significant figures).

Now use this value for t in the general equation for distance with uniform acceleration: x = x + vt + ½at².
Here, x = max height; x = 1m; v = Vv = 5 m/s; a = g = -9.81 m/s²; t = 0.5097 s. Substitute:
x = 1 m + (5 m/s)(0.5097 s) + ½(-9.81 m/s²)(0.5097 s)²
= 2.274 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(.5097 s) = 1.02 s. Use this to find the distance: (8.66 m/s)(1.02 s) = 8.8332 m.

5) If this was phrased properly, Vh = 10 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 = 5 m. (This is 6 m minus the 1 m starting height.) So v = (v² - 2ax) = (0 - 2(-9.81 m/s²)(5 m)) = 9.9 m/s. This is the vertical component of the velocity.

The magnitude of the total velocity is given by (Vh² + Vv²) = (10 m/s)² + (9.9 m/s)² = 14.07 m/s.

6) Since the girls are 8.8322 m apart, use x = vt to find the time of travel (using the horizontal component of the velocity) t = 1.115 s. Now use this in the x = x + vt + ½at² equation. x = 1 m;
v = Vv = 9.9 m/s; a = -9.81 m/s²; t = 1.115 s.
Therefore x = 1 m + (9.9 m/s)(1.115 s) + ½(-9.81 m/s²)(1.115 s)² = 5.94 m.

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