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

$t=\dfrac{-(-45) \pm \sqrt{(-45)^2-4(8)(-80)} }{2(8)}$

$t=\dfrac{45\ \pm\sqrt{4585} }{16}$

$t=7.044\ sec$

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.