# Find the instantaneous velocity

## Homework Statement

a ball is thrown in the air. its height from the ground in metres after t seconds is modelled by h(t) = -5t^2 + 20t + 1. What is the instantaneous velocity of the ball at t=2 seconds?

(f(a+h)-f(a))/h

## The Attempt at a Solution

h(2) = -5(2)2 + 20(2) + 1 = 21
h(2+h) = -5(2+h)2 + 20(2+h) + 1
= -5(2+h)(2+h) + 40 + 20h + 1
= -5(4+2h+2h+h2) + 20h + 41
=-20 -10h – 10h-5h2 + 20h + 41
= -5h2 + 21
lim┬(h→0)⁡〖(f(2+h)-f(2))/h〗 = ((-5h^2+ 21)- 21)/h = (-5h^2)/h = -5h
But then I'm supposed to let h = 0... so the instantaneous velocity is 0. This seems wrong to me? Unless because this is the maximum value, that is when the direction changes, so the velocity is 0?
Thank you for helping!

Dick
Homework Helper

## Homework Statement

a ball is thrown in the air. its height from the ground in metres after t seconds is modelled by h(t) = -5t^2 + 20t + 1. What is the instantaneous velocity of the ball at t=2 seconds?

(f(a+h)-f(a))/h

## The Attempt at a Solution

h(2) = -5(2)2 + 20(2) + 1 = 21
h(2+h) = -5(2+h)2 + 20(2+h) + 1
= -5(2+h)(2+h) + 40 + 20h + 1
= -5(4+2h+2h+h2) + 20h + 41
=-20 -10h – 10h-5h2 + 20h + 41
= -5h2 + 21
lim┬(h→0)⁡〖(f(2+h)-f(2))/h〗 = ((-5h^2+ 21)- 21)/h = (-5h^2)/h = -5h
But then I'm supposed to let h = 0... so the instantaneous velocity is 0. This seems wrong to me? Unless because this is the maximum value, that is when the direction changes, so the velocity is 0?
Thank you for helping!

Yes, it is the maximum value and that's where the direction changes. 0 is correct.

Great, thank you :)