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A rock is dropped from a height of 100 feet. Calculate the time between when the rock was dropped and when it landed. If we choose "down" as positive and ignore air friction, the function is h(t) = 25t^2-81

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The time between the rock’s drop and landing is t = 1.8 s



In the given equation: h(t) = 25t^2 - 81, it is dropped from a height, h, and the variable, t, represents the total time. We want to solve for t.



The height when the rock lands will be 0 m.
Plug 0 into the equation for h and
(0)t = 25t^2 - 81
0 = 25t^2 - 81
81 = 25t^2
Divide both sides by 25 to isolate the variable, t;
81/25 = t^2
Take the square root from both sides to cancel out the exponent;
√(81/25) = √(t^2)
Note that 81 has a sqrt of 9 and 25 has a sqrt of 5. We can also ignore the negative from the ± since time is positive.
9/5 = t
t = 1.8 s
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