Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.
y = -16x2 + 212x + 139​

Answer 1

Answer:

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

Answer 2

Answer:

rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.

y = -16x2 + 212x + 139