- #1

bluemyner

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- Homework Statement
- A ball is thrown from a height of 30.0m with a speed of 10.0m/s at an angle of 23.0 above the horizontal. How fast is the ball going when it hits the ground?

- Relevant Equations
- Vf^2 = Vi^2 + 2a(Delta X)

Delta Y = Vit + 1/2at^2

Vx = 10cos(23) = 9.21m/s

Vy = 10sin(23) = 3.91m/s

-30 = 3.91t + 1/2(-9.8)t^2

0 = t(3.91 - 4.9t) + 30

4.9t = 33.91

t = 6.92s

Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2

Delta X = 298.38m

Vf^2 = 9.21 + 2(9.8)(298.38)

Vf^2 = 5857.458

Vf = 76.53m/s

Vy = 10sin(23) = 3.91m/s

-30 = 3.91t + 1/2(-9.8)t^2

0 = t(3.91 - 4.9t) + 30

4.9t = 33.91

t = 6.92s

Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2

Delta X = 298.38m

Vf^2 = 9.21 + 2(9.8)(298.38)

Vf^2 = 5857.458

Vf = 76.53m/s