Question #a91b2

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Answer 1 · 2016-10-17T09:08:41

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

Differentiating w.r.to t we get the upward velocity $v (t) ↑ ⏐$ $v(t)uparrow$ after t sec.

So

$v (t) ↑ ⏐ ⏐ ⏐ = \frac{d H (t)}{d t} = \frac{d}{d t} (83 t - 16 t^{2})$ $v(t)uparrow=(dH(t))/(dt)=d/(dt)(83t-16t^2)$

$=>v(t)uparrow=83-32t......(1)$

when the ball will be traveling downward at 41.5 feet per second,then
$v(t)uparrow=-41.5" ft/s"$

Inserting this value in (1) we get

$v(t)uparrow=83-32t$

$=>-41.5=83-32t$

$=>t=124.5/32=3.89s$

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