# Projectile motion - maximum height

• Jun 5th 2009, 02:52 PM
peekay
Projectile motion - maximum height
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)

Appreciate any help.
• Jun 5th 2009, 04:43 PM
mr fantastic
Quote:

Originally Posted by peekay
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)

Appreciate any help.

I'd start by calculating the average velocity between the times t = a and t = a + h and then take the limit as h --> 0.
• Jun 6th 2009, 05:17 AM
peekay
I don't know what the variable a and h are referring to (Worried),

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?
• Jun 6th 2009, 05:32 AM
stapel
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. (Wink)
• Jun 6th 2009, 06:37 AM
peekay
So

a + h = (40h-.38h^2) + (40a-.38a^2)
a = (40a-.38a^2)

and then just do m = (y2-y1) / (x2-x1) ?
• Jun 8th 2009, 03:24 AM
peekay
Could anyone solve this and then explain step by step how they did it?

Still having trouble with it :(
• Jun 8th 2009, 03:54 AM
yeongil
Quote:

Originally Posted by peekay
So

a + h = (40h-.38h^2) + (40a-.38a^2)
a = (40a-.38a^2)

and then just do m = (y2-y1) / (x2-x1) ?

The line in red is wrong. You have the function

$\displaystyle S(t) = 40t - 0.83t^2$.

(I wanted to use function notation so I added the parentheses and the t.) You plug in a + h for t:

\displaystyle \begin{aligned} S(a + h) &= 40(a + h) - 0.83(a + h)^2 \\ &= 40a + 40h -0.83(a^2 + 2ah + h^2) \\ &= 40a + 40h -0.83a^2 -1.66ah - 0.83h^2 \\ &= -0.83a^2 - 0.83h^2 - 1.66ah + 40a + 40h \end{aligned}

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,

$\displaystyle S(a) = 40a - 0.83a^2$.

Now use the formula for the slope. The two points are (a, S(a)) and (a + h, S(a + h)):

\displaystyle \begin{aligned} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{S(a + h) - S(a)}{(a + h) - a} \\ &= \frac{(-0.83a^2 - 0.83h^2 - 1.66ah + 40a + 40h) - (40a - 0.83a^2)}{h} \\ &= \frac{-0.83h^2 - 1.66ah + 40h}{h} \\ &= \frac{h(-0.83h - 1.66a + 40)}{h} \end{aligned}
$\displaystyle m = -0.83h - 1.66a + 40$

Now take the limit as h -> 0:
$\displaystyle \lim_{h \to 0} (-0.83h - 1.66a + 40) = -1.66a + 40$

For part b, set -1.66a + 40 equal to 0 and solve for a.

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