Question

A firecracker shoots up from a hill 160 feet high with an initial speed of 90 feet per second. Using the formula H(t) = −16t2 + vt + s, approximately how long will it take the firecracker to hit the ground? ...?

Answer 1

Answer : t = 7.044 sec

Explanation :

It is given that,

Velocity,
`v=90\ feet/sec`

distance,
`s=160\ m`

`H(t)=-16\ t^2+ vt+ s`

When firecracker hits the ground hen H(t) = 0

So,
`-16 t^2+90\ t+160=0`

or

`8 t^2-45\ t-80=0`

Solving this quadratic equation, we get

`t=(-(-45) \pm √((-45)^2-4(8)(-80)) )/(2(8))`

`t=(45\ \pm√(4585) )/(16)`

`t=7.044\ sec`

Neglecting negative values as time can't be negative.

Firecracker will take 7.044 sec to hit the ground.

Answer 2

When the firecracker hits the ground, H(t) = 0.0 = -16t^2 + 90t + 160Using the quadratic formula,t = (-90 +- sqrt(90^2 - 4(-16)(160))) / 2(-16)t = -1.42 ort = 7.04
The negative time is extraneous. Therefore,t = 7.04 s.