The height (in meters) of a bullet fired in the air vertically from ground level with an initial velocity 266 m/s is 𝑠(𝑡)=−4.9𝑡^2+266𝑡. Find the bullet's maximum velocity

Question

The height (in meters) of a bullet fired in the air vertically from ground level with an initial velocity 266 m/s is 𝑠(𝑡)=−4.9𝑡^2+266𝑡. Find the bullet’s maximum velocity and maximum height

Maximum velocity:

Maximum height: 

Answer 1

s(t) -4.9t^2 +266t

s'(t) = -9.8t + 266 = 0

t = 266/9.8 = about 27.14 seconds to reach max height

s(27.14) = -4.9(27.14)^2 + 266(27.14)

= -3610 + 7220 = 3610 meters = max height

max velocity would be when it hits the ground, or was already on the ground: when s=0

s'(t) = v(t) = -9.8(0) +266 = 266 meters per second

or

-9.8(54.28) +266 = -531.944+266 = -266 meters per second.

the negative sign indicates it's going down. initially it's going up. speed is the same at its maximum.

But if the question is maximum velocity, then +266 m/sec is the answer at time t = 0, as +266>-266

speed is the absolute value of velocity