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?

Given : 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 [tex]h = -16t^2 + 272t + 1344[/tex].

To find : When will the ball strike the ground?

Solution :

The equation for the ball's height 'h' at time 't' is given by,

[tex]h(t)= -16t^2 + 272t + 1344[/tex]

When the ball strike the ground the height of the ball became zero.

Substitute h=0 in the given equation,

[tex]-16t^2 + 272t + 1344=0[/tex]

Taking 16 common,

[tex]16(-t^2 + 17t + 84)=0[/tex]

or [tex]t^2-17t-84=0[/tex]

Solve by middle term split,

[tex]t^2-21+4t-84=0[/tex]

[tex]t(t-21)+4(t-21)=0[/tex]

[tex](t-21)(t+4)=0[/tex]

[tex]t=21,-4=0[/tex]

Reject t=-4 as time can never be negative.

Therefore, Time taken to reach the ball to the ground is 21 seconds.