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Help me understand! When a ball is thrown up vertically with velocity vo it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity.

When a ball is thrown up vertically with velocity vo it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity.
 

  • Option 1)

    vo√3 
     

  • Option 2)

    3vo 

  • Option 3)

    9vo 

  • Option 4)

    (3/2)vo

 
Answers (1)
254 Views

As we discussed in concept

Potential Energy -

U_{f}-U_{i}= \int_{r_{i}}^{r_{f}}\vec{f}\cdot \vec{ds}

- wherein

U_{f}-final\: potential\: energy

U_{i}-initial \: potential\: energy

f-force

ds-small \: displacement

r_{i}-initial \: position

r_{f}-final\: position

 

 At Maximum height V=0

V_{0}=\sqrt{2gh} ---(i)

At  h1=3h

u^{2}=2g(3h)=6gh

u=\sqrt{6gh} ---(ii)

\frac{V}{u}=\frac{\sqrt{2gh}}{\sqrt{6gh}}

=> u=\sqrt{3}V_{0}


Option 1)

vo√3 
 

Option is correct

Option 2)

3vo 

Option is incorrect

Option 3)

9vo 

Option is incorrect

Option 4)

(3/2)vo

Option is incorrect

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