A ball is launched into the sky at 272 feet per second from the roof of

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?

2 months ago

Solution 1

Guest Guest #3772
2 months ago

Answer:

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

Step-by-step explanation:

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 h = -16t^2 + 272t + 1344.

To find : When will the ball strike the ground?

Solution :

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

h(t)= -16t^2 + 272t + 1344

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

Substitute h=0 in the given equation,

-16t^2 + 272t + 1344=0

Taking 16 common,

16(-t^2 + 17t + 84)=0

or t^2-17t-84=0

Solve by middle term split,

t^2-21+4t-84=0

t(t-21)+4(t-21)=0

(t-21)(t+4)=0

t=21,-4=0

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

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

Solution 2

Guest Guest #3773
2 months ago
The answer is -16(t+4)(t-21)

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