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 19 m/s at an angle 49 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1. What is the maximum height the ball goes above the ground?

2. What is the distance between the two girls?

3. 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 14 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?

4. 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

given
vo = 19 m/s
theta = 49 degrees
h = 1.5 m

1) maximum height reached by th ball, h_max = h + vo^2*sin^2(theta)/(2*g)

= 1.5 + 19^2*sin^2(49)/(2*9.8)

= 12 m <<<<<<<<-----------ANswer

2) distance between the two girls, R = vo^2*sin(2*theta)/g

= 19^2*sin(2*49)/9.8

= 36.5 m <<<<<<<<-----------ANswer

3) from the given data, vox = 14 m/s

let voy is the initial y-component of velocity.

we know,

h_max = h + voy^2/(2*g)

14 = 1.5 + voy^2/(2*9.8)

vo^2 = (14 - 1.5)*2*9.8

voy = sqrt((14 - 1.5)*2*9.8)

= 15.7 m/s

so, initial speed vo = sqrt(vox^2 +voy^2)

= sqrt(14^2 + 15.7^2)

= 21 m/s <<<<<<<<-----------ANswer

4) let t is the time taken to reach the julie.

t = R/vox

= 36.5/14

= 2.607 s

height of the ball when it gets to the julie,

h = 1.5 + voy*t - (1/2)*g*t^2

= 1.5 + 15.7*2.607 - (1/2)*9.8*2.607^2

= 9.1 m <<<<<<<<-----------ANswer