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`        A projectile of mass m is fired with velocity v at an angle theta with the horizontal what is the change in momentum as the projectile rises to the maximum height `
3 months ago

Swamynathanvp
14 Points
```							as vertical component of velocity becomes zero at highest point but horizontal component does not change Initial momentum = m×vFinal momentum = m×vcos$\Theta$Change in momentum = final - initial momentum = mv(cos$\Theta$ -1)
```
3 months ago
Vikas TU
9801 Points
```							Dear At starting point ux = ucosQuy = usinQAt heighest point Vx = ucosQ Vy =0 Change in velocity along x axis = o Change in velocity along  y axis = - usonQNet change in velocity = -usinQ Change in momentum = -musinQ
```
3 months ago
sumit prakash
26 Points
```							at the highest point projectile have only horizontal velocity that is vcos$\Theta$. so change in momentum in x direction, is mvcos$\Theta$-mvcos$\Theta$=0.change in momentum in y direction is 0-mvsin$\Theta$=-mvsin$\Theta$. finally change in momentum is    -mvsin$\Theta$
```
2 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions