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 12 m/s at an angle 35 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)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 13 m/s when it reaches a maximum height of 7 m above the ground.

5)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.)?

1) To find the horizontal component of the ball's velocity, you can use the formula: horizontal component = initial velocity * cos(angle).

Given that the initial speed is 12 m/s and the angle is 35 degrees, you can calculate it as:
Horizontal component = 12 m/s * cos(35 degrees) = 9.828 m/s.

Therefore, the horizontal component of the ball's velocity when it leaves Julie's hand is approximately 9.828 m/s.

2) To find the vertical component of the ball's velocity, you can use the same formula as before, but with the sine function this time: vertical component = initial velocity * sin(angle).

Using the given values:
Vertical component = 12 m/s * sin(35 degrees) = 6.862 m/s.

So, the vertical component of the ball's velocity when it leaves Julie's hand is approximately 6.862 m/s.

3) To find the maximum height the ball goes above the ground, you can use the formula for vertical motion: maximum height = (vertical component)^2 / (2 * gravity), where gravity is approximately 9.8 m/s^2.

Plugging in the values:
Maximum height = (6.862 m/s)^2 / (2 * 9.8 m/s^2) = 2.8316 m.

Therefore, the maximum height the ball reaches above the ground is approximately 2.8316 meters.

4) To find the distance between the two girls, we need to consider the horizontal component of the ball's motion.

The time taken for the ball to reach Sarah can be found using the formula: time = distance / horizontal component.

Given that Sarah catches the ball 1.5 meters above the ground and the horizontal component is 9.828 m/s, you can calculate it as:
Time = 1.5 m / 9.828 m/s = 0.1528 s.

Now, to find the distance between the two girls, you can use the formula: distance = horizontal component * time.

Distance = 9.828 m/s * 0.1528 s = 1.5039 m.

Therefore, the distance between the two girls is approximately 1.5039 meters.

5) To find the speed of the ball when it leaves Sarah's hand, we can use the formula for vertical motion: speed = sqrt(2 * gravity * height).

Given that the maximum height is 7 m and gravity is approximately 9.8 m/s^2, you can calculate it as:
Speed = sqrt(2 * 9.8 m/s^2 * 7 m) = 19.4 m/s.

Therefore, the speed of the ball when it leaves Sarah's hand is approximately 19.4 m/s.

6) To find how high above the ground the ball will be when it gets to Julie, we can use the same formula for vertical motion: maximum height = (vertical component)^2 / (2 * gravity).

Given that the maximum height is 7 m and gravity is approximately 9.8 m/s^2, you can calculate it as:
7 m = (vertical component)^2 / (2 * 9.8 m/s^2).

Simplifying, you can find that:
(vertical component)^2 = 7 m * 2 * 9.8 m/s^2.

Taking the square root of both sides, you can find that:
Vertical component = sqrt(7 m * 2 * 9.8 m/s^2) = 9.144 m/s.

Therefore, the ball will be approximately 9.144 meters above the ground when it gets to Julie.

1) To find the horizontal component of the ball's velocity when it leaves Julie's hand, we can use basic trigonometry. The horizontal component is given by the formula: horizontal component = initial velocity × cos(angle). In this case, the initial velocity is 12 m/s and the angle is 35 degrees. So, horizontal component = 12 m/s × cos(35 degrees).

2) To find the vertical component of the ball's velocity when it leaves Julie's hand, we can again use trigonometry. The vertical component is given by the formula: vertical component = initial velocity × sin(angle). In this case, the initial velocity is 12 m/s and the angle is 35 degrees. So, vertical component = 12 m/s × sin(35 degrees).

3) To find the maximum height the ball goes above the ground, we can use the equations of motion. The formula to determine the maximum height reached by an object in projectile motion without air resistance is: maximum height = (vertical component)^2 / (2 × acceleration due to gravity). In this case, the vertical component is calculated in step 2, and the acceleration due to gravity is approximately 9.8 m/s^2.

4) To find the distance between the two girls, we need to consider the horizontal motion of the ball. The time of flight can be found using the formula: time of flight = (2 × vertical component) / (acceleration due to gravity). Then, the horizontal distance is given by the formula: horizontal distance = horizontal component × time of flight. In this case, the horizontal component is calculated in step 1.

5) To find the speed of the ball when it leaves Sarah's hand, we need to consider the maximum height reached by the ball. The potential energy at this height is equal to the kinetic energy at the initial height. So, we can use the equation: initial potential energy = final kinetic energy. The initial potential energy is given by: initial potential energy = mass × acceleration due to gravity × initial height. And the final kinetic energy is given by: final kinetic energy = 0.5 × mass × final velocity^2. From these equations, we can solve for the final velocity.

6) To find how high above the ground the ball will be when it gets to Julie, we can use the equations of motion once again. Since the maximum height reached is given, we know that the vertical component of the velocity at that height is 0 m/s. So, we can use the equation: final vertical component = initial vertical component + (acceleration due to gravity × time). From this equation, we can solve for the time needed for the ball to reach Julie. Then, we can use the equation: final height = initial height + (initial vertical component × time) + (0.5 × acceleration due to gravity × time^2).

Vo = 12m/s @ 35 deg.

1. Xo = hor. = 12cos35 = 9.83m/s.

2. Yo = ver. = 12sin35 = 6.88m/s.

3. h = 1.5 + (Vf^2 - Yo^2) / 2g,
h=1.5 + (0 - (6.88)^2) / -19.6 = 3.92m.

4. t(up) = (Vf - Yo) / g,
t(up) = (0 - 6.88) / -9.8 = 0.70s.
t(dn) = t(up) = 0.70s.
T = t(up) + t(dn) = 0.70 + 0.70 = 1.4s
= Time in flight.
d = Xo * T = 9.83 * 1.4 = 13.8m between
the girls.