Question

1. 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 58 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

a. 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 21 m above the ground. What is the speed of the ball when it leaves Sarah's hand?

b. How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)

2. A soccer ball is kicked with an initial horizontal velocity of 18 m/s and an initial vertical velocity of 20 m/s.

a. What is the initial speed of the ball?

b. What is the initial angle ? of the ball with respect to the ground?

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

d. How far from where it was kicked will the ball land?

e. What is the speed of the ball 2.9 seconds after it was kicked?

f. How high above the ground is the ball 2.9 seconds after it is kicked?

Answer 1

a) H = v^2*sin^2(58)/2g

21-1.5 = v^2*sin^2(58)/19.6

V = 23.05m/s

b) H = h + vo^2*sin^58/2g

H = 1.5 + 21*21*sin58*sin58/19.6

H = 17.68 m

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2)

a) v = sqrt(Vx^2+Vy^2)

v = sqrt(18^2+20^2) = 26.9 m/s

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b) tan theta = Vy/Vx

theta = tan^-1(20/18) = 48 degrees

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c) H = Vy^2/2g = 20*20/19.6 = 20.48 m

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d) R = v^2*sin(2*48)/g

R = 73.43 m

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e) V1y = vy - gt = 20-(9.8*2.9) = -8.42 m/s

speed v2 = sqrt(V1y^2+vx^2) = 19.87 m/s

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f) y = vy*t - 0.5*g*t^2

y = 16.791 m