Question

Question #a91b2

Answer 1

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(t)uparrow` after t sec.

So

`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.