When a ball is thrown up into the air, it makes the shape of a parabola. The equ

Question

When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -5t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in meters/sec and “k” is the initial height in meters (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 meters/sec) and any initial height (in meters) 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 of your results explaining them in the context of the original problem.
Please make sure that your answers make sense!

Explanation / Answer

Here is you answer: 1) Take v = 15 m/s and k = 100 m So we have S = -10t^2 +15t +100. 2) When t = 1, S = -10(1)^2 +15(1) + 100 = -10 +15 +100 = 105 meters. When t = 3, S = -10(3)^2 + 15(3) + 100 = -90 + 45 +100 = 55 meters. 3) We throw a ball from the top of a building of height 100 m with an initial velocity of 15 m/s. After 1 second, the height of the ball is 105 meters which means it is 5 meters above the building. After 3 second, the height 55 meters which means it has already started moving in the downward direction and is below the top of the building. Hope this helps.

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