Question

A projectile is fired at an angle of ${30}^0$ to the horizontal such that the vertical component of its initial velocity is $80m{s}^{-1}$. Its time of flight is T. Its velocity at $t=\frac{T}{4}$ has a magnitude of nearly.

Answer 1

Given

$u_y=80m/s$

$\theta =30$

$u_y=u\sin \theta$

$u=80\times 2=160m/s$

At the most height, $v_y=0$

$v_y=u_y+\frac{aT}{2}$

$0=80+\frac{10\times T}{2}$

$T=16s$

Velocity after time $\frac{T}{4}$

$v_y=80-\frac{10\times 4}{4}$

$v_y=40m/s$

Horizontal component

$u_x=u\cos \theta$

$u_x=80\sqrt{3}m/s$

$V=\sqrt{u_x^2+v_y^2}$

$V=\sqrt{(80\sqrt{3}{)}^2+(40{)}^2}$

$V=144.22m/s$