Question

A particle is thrown from ground with speed u at an
angle with horizontal. The average velocity of the
particle in the time interval t = 0 to t=where T
is time of flight of projectile, is​

Answer 1

answer : $\frac{u}{2}\sqrt{1+3cos^2\theta}$

explanation : velocity after time t is given as, $v=ucos\theta\hat{i}+(usin\theta-gt)\hat{j}$

using formula to find average velocity,

$v_{av}=\frac{\int{v(t)}\,dt}{\int{dt}}$

so, average velocity between 0 to T/2.

$a_{av}=\frac{\int\limits^{T/2}_0{(ucos\theta\hat{i}+(usin\theta-gt)\hat{j})}\,dt}{\int\limits^{T/2}_0{dt}}$

= $\frac{[ucos\theta.t\hat{i}+\left(usin\theta.t-\frac{gt^2}{2}\right)\hat{j}]^{T/2}}{T/2}$

= [ucosθ(T/2)i + {usinθ(T/2)- gT²/8}j]/(T/2)

= ucosθi + (usinθ - gT/4)j

now, putting, T = 2usinθ/g

= ucosθi + {usinθ-g/4(2usinθ)/g}j

= ucosθi + {usinθ-(usinθ)/2}j

= ucosθi +{u(sinθ - sinθ/2)}j

= ucosθi + u(sinθ)/2 j

so, magnitude of average velocity, $|a_{av}|=u\sqrt{cos^2\theta+sin^2\theta/4}$

= $\frac{u}{2}\sqrt{4cos^2\theta+sin^2\theta}$

= $\frac{u}{2}\sqrt{1+3cos^2\theta}$

[using sin²θ + cos²θ = 1]

Answer 2

$\huge\bold\purple{Answer:-}$

-2vsintheta/T................