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 21 m/s at an angle 59 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

a. What is the vertical component of the ball’s velocity right before Sarah catches it?

b. What is the distance between the two girls?

c. After catching the ball, Sarah 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 1.5 meters above the ground, reaches a maximum height of 21 m above the ground, and takes 2.835 s to get directly over Julie's head.

What is the speed of the ball when it leaves Sarah's hand?

d. How high above the ground will the ball be when it gets to Julie?

Answer 1

a)

Initial Horizontal and vertical components of velocities are

V_ox=21Cos59 =10.8158 m/s

V_oy=21Sin59 =18.0005 m/s

From Kinematic equation

V_fy² =V_oy²-2g(y-y_o)

V_fy²=18.0005²-2*9.81*(1.5-1.5)

V_fy=-18.0005 m/s =-18 m/s (approx)

b)

From Kinematic equation

Y=Y_o+V_oyt-(1/2)gt²

1.5 = 1.5 +18.0005*t-(1/2)(9.81)t²

t=[2*18.0005/9.81] =3.67 s

Horizontal distance

X=V_oxt =10.8158*3.67 =39.7 m

c)

initial Horizontal Velocity

V_ox=X/t =39.7/2.835 =14.0014 m/s

From kinematic equation

V_fy²=V_oy²-2g(Y-Y_o)

at maximum height V_fy=0

0=V_oy²-2*9.81*(21-1.5)

V_oy = 19.56 m/s

Speed of the ball as it leaves sarah's hand is

V_o=sqrt[14.0014²+19.56²] =24.05 m/s

d)

From

Y=Y_o+V_oyt-(1/2)gt²

Y=1.5+19.56*2.835-(1/2)(9.81)(2.835)²

Y=17.53 m