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`        a particle is projected from the ground with an initial an initial speed at an angle theata with the horizontal. The average velocity between the point of projection and the highest point of trajectory is`
11 months ago

```							Vi = VcosQ  i + VsinQ jVf = VcosQ i Vav = (Vi + Vf )/2Vav = V/2 (sqrt (2cosQ)2 + (2SinQ)2 )=> V/2 (sqrt ( 4cos2Q + sin2Q) => V/2 (sqrt (3cos2Q +1)
```
11 months ago
```							you could use integration…the velocity in the x direction is always constant. so the average velocity in the x direction is also the constant velocity with which the body travels i.e u(cos theta)…but the velocity in the y direction is a little trickieryou have to use the average formula for integration…it goes like this…lets assume that the velocity in the y direction is a function of y…let that bevelocity in the y direction = f(y)(i am assuming that the actual function would be easy to find out…using the basic laws of motion )then integrate the f(y) with the limits from y=0 to y=(maximum height attained by a projectile ) (the formula is easily found on the internet)after you find out the integral, divide the answer by the maximum height…now you have found the average velocity in the y directionhowever you already know the average velocity in the x direction…using them as components find out the average velocity as a total([velocity in the y direction]^2 + [velocity in the x direction]^2)^(1/2)i thank that you should get your answer as({u^2(1+3[cos theta]^2)}^(1/2) )*(1/2)
```
11 months ago
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