If a projectile is fired upward from the surface of the moon with an initial velocity of 40 m/s, its height, in metres, after t seconds is given by s=40t-.83t^2
a) Estimate the instantaneous velocity when t = a, (without using calculus)
b) Using your answer in part a, determine the maximum height.
Appreciate any help.
I don't know what the variable a and h are referring to ,
There were actually two questions before question a and b, and they were to find the average velocity between 1 and 3 seconds which is 38.705 m/s and to also find the instantaneous velocity at 2 seconds, which is 38.4763 m/s. Am I suppose to do something with the instantaneous velocity when t = 2?
In the earlier parts, they gave you a = 1 and a + h = 3 (so h = 2). Here you have a and a + h, so do the exact same steps as you did before.
The only difference will be that you won't be able to simplify as much, because you're working with a variable instead of which numbers.
(I wanted to use function notation so I added the parentheses and the t.) You plug in a + h for t:
Also, the line underneath the red one above isn't completely correct. You're not really saying that a equals (40a-.38a^2), but this is what the expression gives you when you plug in a. In function notation,
Now use the formula for the slope. The two points are (a, S(a)) and (a + h, S(a + h)):
Now take the limit as h -> 0:
For part b, set -1.66a + 40 equal to 0 and solve for a.