Question

A projectile launched straight up in the air has its height in seconds after launch given by the function

F(t)=−16t ^2 +382t+7. Find its velocity 9.4 seconds after it is launched.

A) 325 m/s
B) 342 m/s
C) 371 m/s
D) 396 m/s

Answer 1

The velocity of the projectile 9.4 seconds after it is launched is 342 m/s.

Step-by-step explanation:

To find the velocity of the projectile 9.4 seconds after it is launched, we need to find the derivative of the height function F(t) with respect to time. The derivative of F(t) gives us the velocity function, which represents the rate of change of height with respect to time. Taking the derivative of F(t)=−16t^2+382t+7, we get F'(t)=-32t+382. Substituting t=9.4 into the velocity function, we find that the velocity of the projectile 9.4 seconds after it is launched is 342 m/s. Therefore, the correct answer is B) 342 m/s.