Question

A projectile is thrown with velocity V at an angle with the horizontal . When the projectile is at height equal to half of the maximum height then the vertical component of the velocity of projectile is-

Answer 1

A projectile is thrown with velocity v at an angle $\theta$ with horizontal.

then, $v_x=vcos\theta ,v_y=vcos\theta$

we have to find vertical component of velocity of the projectile at height equal to half of the maximum height.

we know, $H_{max}=\frac{u^2{sin}^2\theta }{2g}$

here u is intial velocity of projectile. but here given initial velocity is v

so,$H_{max}=\frac{v^2{sin}^2\theta }{2g}$

now, time taken to reach half of maximum height, t

$Y=v_{y}t+\frac{1}{2}{a}_y{t}^2$

$\frac{v^2{sin}^2\theta }{2g}=vsin\theta .t-5t^2$

$200t^2-40vsin\theta .t+v^2{sin}^2\theta =0$

$t=\frac{(\sqrt{2}\pm 1)vsin\theta }{10\sqrt{2}}$

${v_y}^{{}^{\prime }}=v_y+a_{y}t$

$=vsin\theta -\frac{(\sqrt{2}-1)vsin\theta }{\sqrt{2}}$

$=\frac{vsin\theta }{\sqrt{2}}$