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Grade: 11
        
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
4 months ago

Answers : (2)

Vikas TU
9447 Points
							
Vi = VcosQ  i + VsinQ j
Vf = VcosQ i 
Vav = (Vi + Vf )/2
Vav = V/2 (sqrt (2cosQ)2 + (2SinQ)2 )
=> V/2 (sqrt ( 4cos2Q + sin2Q) 
=> V/2 (sqrt (3cos2Q +1)
4 months ago
Khimraj
3008 Points
							

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 trickier

you 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 be

velocity 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 direction

however 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)

3 months ago
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