Question

The angle which the velocity vector of a projectile, thrown with a velocity $v$ at an angle $\theta$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is

• A. $\theta$
• B. $ta{n}^{-1}\left(\frac{\theta }{t}\right)$
• C. $ta{n}^{-1}\left(\frac{vcos\theta }{vsin\theta -gt}\right)$
• D. $ta{n}^{-1}\left(\frac{vsin\theta -gt}{vcos\theta }\right)$

Answer 1

The correct option is D $ta{n}^{-1}\left(\frac{vsin\theta -gt}{vcos\theta }\right)$

Horizontal velocity of projectile after time t $=vcos\theta$
Vertical velocity of projectile after time t$=vsin\theta -gt$
Angle at any time t:

$\tan \varphi =\frac{v_v}{v_h}=\frac{vsin\theta -gt}{vcos\theta }\varphi ={tan}^{-1}\frac{vsin\theta -gt}{vcos\theta }$