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? ...?
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.
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.