Question

A ball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. The equation for the ball’s height h at time t seconds is h = -16t^2 + 272t + 1344. When will the ball strike the ground?

Answer 1

Hello @Wpisd3039,

How are you doing? In this case,
The ball will strike the ground when h=0.-16t²+272t+1344=0t₁,₂=(-272+-√(272²+4*16*1344))/2*(-16)t₁,₂=(-272+-400)/(-32)t₁=(-272+400)/(-32)=-4 or t₂=(-272-400)/(-32)=21Time cannot be negative value so t₂ is solution. The ball will strike the ground in 21 seconds.

I hope I helped, Contact me if you need more assistance.

Thank you,
Darian D.<span />