The height of a falling object

Question

The height of a falling object is given by y(t) = 89 − 4.9t 2 , with height y in meters and time t in seconds.

(a) compute y' (a).

(b) When will the velocity be −30 m/s?

Answer 1

I assume the formula for y(t) is supposed to read 89-4.9t², not t 2.

For a, you just need to find the derivative of the given formula for y(t). This is pretty straightforward:

dy/dt[89-4.9t²] = 0 - 4.9*2*t = -9.8t. We calculate this derivative by using the power rule for each part of the equation. The derivative of 89 is 0, because it is a constant. The derivative of -4.9t² is found by multiplying -4.9t by the exponent (2) and then subtracting 1 from the exponent. This is the power rule. Doing that, we get -4.9 * 2 * t = -9.8t.

For b., you need to understand that velocity is the derivative of position. Also understand that acceleration is the derivative of velocity (although that isn't relevant to this problem). Therefore, the formula for velocity is the derivative of our height formula (which is our "position" formula). We found this in part a:

v(t) = -9.8t.

To find when the velocity will be -30 m/s, we just need to set our velocity formula equal to -30:

v(t) = -9.8t = 30.

Now it's just simple algebra to solve for t!

t = 30/-9.8 = -3.061 m/s (rounded 3 decimal places, don't forget your units).