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 17 m/s at an angle 34 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 right before Sarah catches it? 2) what is the vertical component of the ball's velocity right before Sarah catches it? 3) what is the time the ball is in the air? 4) what is the distance between the two girls?

1) no horizontal acceleration

... v = 17 m/s * cos(34º)

2) same vertical velocity as at launch, except in opposite direction
... v = 17 m/s * sin(34º) ... downward

3) the flight time is twice the initial vertical velocity, divided by gravitational acceleration

4) the distance is the (constant) horizontal velocity, multiplied by the flight time

To solve this problem, we can use projectile motion equations. Projectile motion occurs when an object is launched into the air and moves along a curved path under the influence of gravity. In this case, the ball thrown by Julie follows a curved path until it reaches Sarah.

Let's break down each question and solve them individually:

1) To find the horizontal component of the ball's velocity right before Sarah catches it, we need to determine how fast the ball is moving horizontally. In projectile motion, the horizontal velocity remains constant throughout the motion.

The horizontal component of the initial velocity can be found using the formula:
vx = v * cosθ, where vx is the horizontal component of velocity, v is the initial speed, and θ is the angle with respect to the horizontal.

In this case, the initial speed (v) is 17 m/s, and the angle (θ) is 34 degrees. Substituting these values into the formula:
vx = 17 m/s * cos(34°) ≈ 14.03 m/s

Therefore, the horizontal component of the ball's velocity right before Sarah catches it is approximately 14.03 m/s.

2) To find the vertical component of the ball's velocity right before Sarah catches it, we need to determine how fast the ball is moving vertically. In projectile motion, the vertical component of velocity changes due to the influence of gravity.

The vertical component of the initial velocity can be found using the formula:
vy = v * sinθ, where vy is the vertical component of velocity, v is the initial speed, and θ is the angle with respect to the horizontal.

Substituting the known values:
vy = 17 m/s * sin(34°) ≈ 9.69 m/s

Therefore, the vertical component of the ball's velocity right before Sarah catches it is approximately 9.69 m/s.

3) To find the time the ball is in the air, we can use the formula for the time of flight (t):
t = (2 * vy) / g, where t is the time of flight, vy is the vertical component of velocity, and g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the known values:
t = (2 * 9.69 m/s) / 9.8 m/s² ≈ 1.97 seconds

Therefore, the time the ball is in the air is approximately 1.97 seconds.

4) To find the distance between the two girls, we need to determine the horizontal distance covered by the ball.

The horizontal distance can be calculated using the formula:
distance = vx * t, where distance is the horizontal distance, vx is the horizontal component of velocity, and t is the time of flight.

Substituting the known values:
distance = 14.03 m/s * 1.97 seconds ≈ 27.59 meters

Therefore, the distance between the two girls is approximately 27.59 meters.

To find the answers, we can use the equations of motion for projectile motion.

1) The horizontal component of the ball's velocity remains constant throughout the motion. The initial horizontal velocity can be found using the equation:

vx = initial velocity * cos(angle)

Given that the initial speed is 17 m/s and the angle is 34 degrees, we can calculate the horizontal component:

vx = 17 m/s * cos(34 degrees)

Using a calculator, we find that vx is approximately 14.07 m/s.

2) The vertical component of the ball's velocity changes due to the effect of gravity. To find the vertical component right before Sarah catches the ball, we need to find the time it takes for the ball to reach Sarah's hand.

The vertical motion can be analyzed using the equation:

vf = vi + at

At the highest point of the motion, the vertical component of the final velocity (vf) is zero. The acceleration due to gravity (a) is -9.8 m/s² (taking downward as negative). The initial vertical velocity (vi) can be found using the equation:

vi = initial velocity * sin(angle)

vi = 17 m/s * sin(34 degrees)

Using a calculator, we find that vi is approximately 9.12 m/s.

Using these values, we can solve for the time using the equation:

0 = 9.12 m/s + (-9.8 m/s²) * t

Solving for t, we find that t is approximately 0.93 seconds.

3) The total time the ball is in the air is twice the time it takes to reach the highest point (t_total = 2t). Therefore, the time the ball is in the air is approximately 1.86 seconds.

4) To find the horizontal distance between the two girls, we can use the equation:

distance = horizontal velocity * time

Given that the horizontal velocity (vx) is 14.07 m/s and the time (t_total) is 1.86 seconds, we can calculate the distance:

distance = 14.07 m/s * 1.86 s

Using a calculator, we find that the distance is approximately 26.14 meters.