Question

PLEASE ANSWER QUICKLY :) A rocket is launched vertically from the ground with an initial velocity of 64 ft/sec. Write a quadratic function that shows the height, in feet, of the rocket t seconds after it was launched. You don't need to show your work.

Answer 1

Answer:

The equation for the height, in ft, is:

$h(t)=-16.085\cdot t^2+64\cdot t$

Step-by-step explanation:

The height can be expressed as an integration of the speed:

$h(t)-h(0)=h(t)=\int_{0}^t v(t)\,dt$

Also, the speed can be expressed as an integration of the acceleration:

$v(t)-v(0)=v(t)-64=\int_0^ta(t)\,dt=-g\cdot t=-32.17t\\\\v(t)=64-32.17t \;[ft/s]$

Then, if we go to the height equation, we have:

$h(t)=\int_{0}^t v(t)\,dt=64t-32.17\cdot \dfrac{t^2}{2}=64t-16.085t^2$

The equation for the height is:

$h(t)=-16.085\cdot t^2+64\cdot t$