Question

A projectile is projected from ground with initial velocity $\overrightarrow{u}=u_0\hat{i}+v_0\hat{j}.$ If acceleration due to gravity (g) is along the negative y-direction then find maximum displacement in x-direction.

• A. $\frac{u_0^2}{2g}$
• B. $\frac{{2u}_0{v}_0}{g}$
• C. $\frac{v_0^2}{2g}$
• D. $\frac{{4u}_0{v}_0}{g}$

Answer 1

The correct option is B $\frac{{2u}_0{v}_0}{g}$

The correct option is B.

Given,

$\overrightarrow{U}=u_0\hat{i}+v_0\hat{j}$

$u_0=ucos\theta .......1$

$v_0=vsin\theta ........2$

Since,

$R=\frac{u^{2}2sin2\theta }{g}$

$R=\frac{u^{2}2(sin\theta .cos\theta )}{g}...........3$

Substituting the value of equation 1,2 in eqyuation 3 then we get,

$R=\frac{2u_0{v}_0}{g}$