What is the instantaneous velocity of the ball at t=2 seconds if a ball is thrown in the air and its height from the ground in meters after t seconds is modeled by #h(t)=-5t2+20t+1#?

Question

Ezekiel Alexander · Accepted Answer

#h=21m#.

It is sufficient to substitue #t=2# in that equation: #h(2)=-5*2^2+20*2+1=21m#.

Adam Adkins · Answer

undefined To find the instantaneous velocity of the ball at $ t = 2 $ seconds, we need to find the derivative of the height function $ h(t) $ with respect to time $ t $, which gives us the velocity function. Then, we can evaluate this velocity function at $ t = 2 $ seconds to find the instantaneous velocity.

The height function is given by $ h(t) = -5t^2 + 20t + 1 $.

The velocity function, which is the derivative of the height function, is given by $ v(t) = h'(t) $.

Taking the derivative of $ h(t) $ with respect to $ t $, we get:

$$ h'(t) = \frac{dh}{dt} = -10t + 20 $$

Now, we can evaluate $ v(t) $ at $ t = 2 $ seconds:

$$ v(2) = -10(2) + 20 $$
$$ v(2) = -20 + 20 $$
$$ v(2) = 0 $$

So, the instantaneous velocity of the ball at $ t = 2 $ seconds is $ 0 $ meters per second.