Question

The angle of projection with the horizontal at which projectile is projected is

• A. ${\tan }^{-1}\frac{3}{4}$
• B. ${\tan }^{-1}\frac{4}{3}$
• C. ${\sin }^{-1}\frac{3}{4}$
• D. ${\cos }^{-1}\frac{3}{4}$

Answer 1

The correct option is A ${\tan }^{-1}\frac{3}{4}$

as for vertical direction equation of motion

Y =v sin $\theta$t -$\frac{1}{2}g{t}^2$

and for horizontal direction

x=Vcos $\theta$t

now comparing with given equation

y=6t-1$10t^2$

x=8t

So, V cos $\theta$=8

V sin $\theta$=6

So, divide them

$\frac{Sin\theta }{cos\theta }=\frac{6}{8}$

So,tan $theta$ =$\frac{3}{4}$

$\theta =ta{n}^{-1}\frac{3}{4}$

As in this question we know general equation of projectiles

in y direction

in x direction

now on comparing with the given equation as given in the picture

we get $\tan \theta =3/4$

so $\theta =ta{n}^{-1}\left(3/4\right)$