# Finding the magnitude of velocity vector

## Homework Statement

A cricketer throws a 147g ball by exerting a force of 7.5N for 0.84s. If the launch angle is 39 degrees from horizontal, calculate the range of the ball.

## The Attempt at a Solution

I have the range equation and I just need to find the magnitude of the velocity vector but I'm not sure how I'd find that with the given quantities?

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What is the answer given in solutions section?

SammyS
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## Homework Statement

A cricketer throws a 147g ball by exerting a force of 7.5N for 0.84s. If the launch angle is 39 degrees from horizontal, calculate the range of the ball.

## The Attempt at a Solution

I have the range equation and I just need to find the magnitude of the velocity vector but I'm not sure how I'd find that with the given quantities?
Can you determine the acceleration ?

NoahCygnus
I thought it was a=f/m = 51.02m/s^2 and then v=at for 42.86 ms^-1 I was going to use a particular formula for range

The solution in the back of the mock exam says f=ma rearranged for m x deltav/deltat

Which is then rearranged for deltav = Fdeltat / m = 42.8 ms

Trigonometry to then find the Vx and Vy components which can then be plugged into t = v-u/a giving us a time value. This completes the vertical plane suvat.

Horizontal 'range' was simply given as displacement s=vt: 42.8(cos39) x 6.78 = 225.52m

When would this equation for range be applicable? It's defined as 'the displacement in the x axis when the y axis displacement is 0.' I imagine there is still some displacement for this scenario even though it doesn't end in the y axis? Making this equation irrelevant?