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A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground? 

Guest Mar 5, 2017

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 #2
avatar+6705 
+5

A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground?

 

\(mgh=\frac{1}{2}mv^2\)

 

\(\large h=\frac{v^2}{2g}\)

 

\(\Large h=\frac{\frac{12^2m^2}{s^2}}{2\times 9.81\frac{m}{s^2}}\)

 

\(\large h=7.339m\)

 

laugh !

asinus  Mar 5, 2017
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 #1
avatar+4161 
+6

I think I can do this problem, but I'm not the best at these so it could be wrong.

 

Using 9.8 m/s2 as the approximation for gravity:

acceleration function: →  h''(t) = -9.8

velocity function: →         h'(t) = -9.8t + 12

position function: →         h(t) = -4.9t2 + 12t

 

I didn't explain all the steps I did there because I don't know if my way of doing it is the best way; it's just the only way I know how to do it. So I don't want to confuse you any if my way is not the way you've been taught to do it.

 

We need to find the values of t where the velocity, h'(t) = 0. The velocity will be zero right at the max height.

 

0 = -9.8t + 12

-12/-9.8 = t

60/49 = t

 

So at t = 60/49, the velocity is 0. So let's find the position, h(t), when t = 60/49.

 

h(60/49) = -4.9(60/49)2 + 12(60/49)

h(60/49) = -4.9(3600/2401) + 720/49

h(60/49) = -360/49 + 720/49

h(60/49) = 360/49 ≈ 7.347 meters

hectictar  Mar 5, 2017
 #2
avatar+6705 
+5
Best Answer

A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground?

 

\(mgh=\frac{1}{2}mv^2\)

 

\(\large h=\frac{v^2}{2g}\)

 

\(\Large h=\frac{\frac{12^2m^2}{s^2}}{2\times 9.81\frac{m}{s^2}}\)

 

\(\large h=7.339m\)

 

laugh !

asinus  Mar 5, 2017

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