A water balloon is launched upward with an initial velocity of $40 ft/s$. the height h at time $t$ is given $h(t)= 16t^2+40t$
a. find the avg. velocity of the balloon between $t=1$ and $t=2$ seconds
b. find the instantaneous velocity at $t=1$sec
I'm confused how to find the answers when given the formula for $h$..do you take the derivative and substitute $t$? If anyone can let me know how to set up each part of the problem that would be extremely helpful
Best Answer
a) $v_{ave} = \dfrac{h(2) - h(1)}{2 - 1} = \dfrac{16\cdot 2^2 + 40\cdot 2 - 16\cdot 1^2 - 40\cdot 1}{1} = 88 ft/s$.
b) $v_{int} = h'(1) = 32t + 40 |_{t = 1} = 32\cdot 1 + 40 = 72 ft/s$