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 = -16t2 + 272t + 1344. When will the ball strike the ground?

Answer 1

The ball will strike the ground when h=0.

-16t²+272t+1344=0

t₁,₂=(-272+-√(272²+4*16*1344))/2*(-16)

t₁,₂=(-272+-400)/(-32)

t₁=(-272+400)/(-32)=-4 or t₂=(-272-400)/(-32)=21

Time cannot be negative value so t₂ is solution. The ball will strike the ground in 21seconds.

I hope this helped!

Explanation:

Answer 2

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