# Question a91b2

##### 1 Answer
Oct 17, 2016

A ball is thrown upward in the air, and its height above the ground after t seconds is H(t)=83⋅t-16t^2# feet.

Differentiating w.r.to t we get the upward velocity $v \left(t\right) \uparrow$ after t sec.

So

$v \left(t\right) \uparrow = \frac{\mathrm{dH} \left(t\right)}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(83 t - 16 {t}^{2}\right)$

$\implies v \left(t\right) \uparrow = 83 - 32 t \ldots \ldots \left(1\right)$

when the ball will be traveling downward at 41.5 feet per second,then
$v \left(t\right) \uparrow = - 41.5 \text{ ft/s}$

Inserting this value in (1) we get

$v \left(t\right) \uparrow = 83 - 32 t$

$\implies - 41.5 = 83 - 32 t$

$\implies t = \frac{124.5}{32} = 3.89 s$

So the ball will be traveling downward at 41.5 feet per second after 3.89s of its projection upward from the ground.