14. A ball is projected upward at time t = 0.0 s from a point on a roof 90 m ab

Physics
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14. A ball is projected upward at time t = 0.0 s from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 80.5m/s The velocity of the ball when it is 89m above the ground is about
May 18th, 2015

We use the formula:  h(t) = -16t^2 + Vit + ho

h(t) = -16t^2 + 80.5t + 90

the ball reaches maximum height at t = 80.5/32 = 2.516 s

plugging back into equation gives us maximum height of 191.25 m

After that, the ball is in free fall under the force of gravity with acceleration of 9.8 m/s

The distance from peak height to 89m is 191.25 - 89 = 102.25 m

102.25 / 9.8 = 10.43 s^2 = 3.23 s

Therefore the velocity is (9.8)(3.23) = 31.66 m/s,  but since it's in the downward direction it is -31.66 m/s


May 18th, 2015

Oops, correction.  d = 1/2gt^2

so, 102.25 = 1/2(9.8)t^2

204.5 = 9.8t^2

t^2 = 204.5/9.8 = 20.87 s

t = 4.57

therefore, velocity = (-9.8)(4.57) = -44.77 m/s

May 18th, 2015

Sorry again, my first formula was wrong because it was in feet instead of meters

h(t) = -4.9t^2 +Vot + ho

= -4.9t^2 + 80.5t + 90

time to reach max height = 80.5/9.8 = 8.2 seconds

plug back in to find max height, calculate and get 420.6 m

420.6 - 89 = 331.6 meters

331.6 = 1/2(9.8)t^2

663.25 = 9.8t^2

t = 8.2 s

velocity = (-9.8)(8.2) = -80.6 m/s

May 18th, 2015

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May 18th, 2015
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May 18th, 2015
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