Calculus help?

Question

Let s(t)=-5t^2+40t+1 describe a toy rocket's motion as it is launched vertically in the air. Determine the maximum height, in metres, that it reaches over the time interval 1<=t<=5

Answer 1

s(t) = - 5t^2 + 40t + 1

The rate of change in displacement is called velocity.

Velocity is distance traveled per time together with the direction of motion.

The SI unit of distance is the meter (m) and the SI unit of time is the second (s).

Derivative of the function

s(t) = - 5t^2 + 40t + 1

v = ds/dt = -10t + 40

Maximum height is where the velocity is zero.

-10t + 40 = 0

-10t = -40

t = 4 sec

The rocket reaches the maximum height  in 4 sec.

1 <= 4 <=5

Substitute the t  value in the function.

 = -5(4)^2 + 40(4) + 1

= -80 + 160 + 1

= 81

Maximum height of toy rocket is 81 meters.