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When a ball is thrown up into the air

# When a ball is thrown up into the air - of your results...

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When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is the initial height in feet (as if you were on top of a tower or building). Make up a scenario where a ball is thrown, shot, etc. into the air. You can choose any initial velocity (in feet/sec) and any initial height (in feet) of the ball, but include them in your written scenario. The ball can leave your hand, the top of a building, etc. so you can use many different values for the initial height. Insert the chosen values for “v” and “k” into the formula listed above. Use the formula to find the height of the ball at any two values of time, t, in seconds that you want. Show your calculations and put units on your final answer! Provide a written summary

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Unformatted text preview: of your results explaining them in the context of the original problem. Consider a ball thrown from a height of 10ft with initial velocity 2 m/s So it means v = 2 m/s , k = 100 ft, So the formula of height at any time t becomes S=-16t 2 +2t + 100 Now consider t = 2 seconds S = -16(2) 2 + 2(2) + 100 S = -64 + 4 + 100 S = 40 ft So after 2 seconds the ball is 40ft above the ground Consider t = 1 seconds S = -16(1) + 2(1) + 100 S = 86 ft above the ground So from above we see that at t = 2 seconds the ball is at 40 ft height while at t =1 second it is 86 ft, since the ball is thrown from 100 ft so its obvious that at some time later it will be at a lower depth Now the height becomes 0 when -16t 2 + 2t+ 100 = 0 Which on solving gives t = 2.56 seconds So at t = 2.56 seconds the ball will reach the ground Here the is graph which shows the trajectory of the ball...
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When a ball is thrown up into the air - of your results...

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