Calculus help!
A cricket jumps in the air with a velocity of 2.6 m/s, its height (m) after 1 second is given by h=2.6t-4.9t^2
A) how fast is the cricket moving after t=0.05 s?
B) what is the average between the start and t=0.05s?
C) it will reach the highest point and start to fall down when the velocity is zero. At what time will this happen?

Question

Calculus help!
A cricket jumps in the air with a velocity of 2.6 m/s, its height (m) after 1 second is given by h=2.6t-4.9t^2
A) how fast is the cricket moving after t=0.05 s?
B) what is the average between the start and t=0.05s?
C) it will reach the highest point and start to fall down when the velocity is zero. At what time will this happen?

anonymous · Answer

For A, we need to take the first derivative of $h$.

anonymous · Answer

2.6-9.8t?

anonymous · Answer

Yes. Now plug t = .05 into that equation to obtain your answer for A.

anonymous · Answer

okay, what about part b?

anonymous · Answer

I believe it is asking for speed, which is $\frac{distance}{time}$. I could be wrong because that question is slightly unclear.