Find the instantaneous velocity

  • Thread starter pbonnie
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  • #1
pbonnie
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Homework Statement


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


Homework Equations


(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!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement


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


Homework Equations


(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.
 
  • #3
pbonnie
92
0
Great, thank you :)
 

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