A rocket is launched from atop a 105 foot cliff with an initial velocity of 156 ft/s . The height of the rocket above the ground at time t is given by h=-16t^2+156t+105. When will the rocket hit the ground after it is launched? Round to the nearest tenth.

A) 4.9 s
B) 9.8 s
C) 0.6 s
D) 10.4 s

Question

A rocket is launched from atop a 105 foot cliff with an initial velocity of 156 ft/s . The height of the rocket above the ground at time t is given by h=-16t^2+156t+105. When will the rocket hit the ground after it is launched? Round to the nearest tenth.

A) 4.9 s
B) 9.8 s
C) 0.6 s
D) 10.4 s

anonymous · Answer

@♂

anonymous · Answer

When the rocket hits the ground, h=0. The equation then becomes$$-16t^2+156t+105=0$$Solve for t.

anonymous · Answer

so 4.9 ? @ospreytriple

anonymous · Answer

What are the two solutions you got for t by solving the quadratic?

anonymous · Answer

$$t = \frac{ 39 }{ 8 } - \frac{ \sqrt{1941} }{ 8 }$$

$$t = \frac{ 39 }{ 8 } + \frac{ \sqrt{1941} }{ 8 }$$

anonymous · Answer

I used the quadratic formula and got t=-0.6 (discard, we can't have negative time)
and t=10.38 which rounds to 10.4

waiting for osprey to finish before posting this so I'm not just spoonfeeding you

anonymous · Answer

i think i did it wrong

anonymous · Answer

thank you! @♂