Question

If a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y = 40t − 16t2. Find the velocity when t = 2 seconds.

Answer 1

Height of the function is given by $y=40t-16t^2$

Require to find the velocity when $t=2$ seconds

To find the velocity, find $\frac{dy}{dt}$ from the given function $y=40t-16t^2$

Now $y=40t-16t^2\Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left [40t-16t^2 \right ]$

$\Rightarrow \frac{dy}{dt}= 40(1)-16(2t)$

$\Rightarrow \frac{dy}{dt}= 40-32t$

So, velocity is given by $v(t)=40-32t$

Now let us find the velocity when $t=2$ seconds

$t=2\Rightarrow v(2)=40-32(2)=40-64=-24$

Therefore, $v(2)=-24 ft/s$