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 of your results explaining them in the context of the original problem.
Please make sure that your answers make sense!
Just follow the directions
Since 16 is used for (1/2)g and not 4.9, this problem is in feet units and not meters.
choose an initial height k, for example 30 feet.
choose an initial speed up v, for example 10 ft/s
then
S = -16 t^2 + 30 t + 30
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Now how high is the ball after 1 second?
How high is it after 2 seconds?
Note that it is perfectly all right for it to be below ground. There is nothing in the stated problem to tell it to stop when it hits the ground. In fact we could solve for when S = 0
16 t^2 - 30 t - 30 = 0
solve that quadratic if you want to know when it hits ground.
Scenario: A ball is thrown from the top of a building with an initial velocity of 30 ft/sec and an initial height of 100 feet.
Using the equation S = -16t^2 + vt + k, where v is the initial velocity and k is the initial height, we can calculate the height of the ball at different times.
Let's find the height of the ball at t = 2 seconds and t = 4 seconds.
For t = 2 seconds:
S = -16(2)^2 + 30(2) + 100
S = -64 + 60 + 100
S = 96 feet
For t = 4 seconds:
S = -16(4)^2 + 30(4) + 100
S = -256 + 120 + 100
S = -36 feet
Summary:
At t = 2 seconds, the height of the ball is 96 feet.
At t = 4 seconds, the height of the ball is -36 feet.
In the context of the problem, these answers make sense. At t = 2 seconds, the ball has reached a point higher than its initial height, indicating that it is ascending. At t = 4 seconds, the ball has reached a height below its initial height, indicating that it is on its way down towards the ground.