For a projectile launched with speed $v_0$ at angle $ heta$, the hori…

Question

For a projectile launched with speed $v_0$ at angle $	heta$, the horizontal range is

Accepted Answer

B. $R=\dfrac{v_0^2\sin2	heta}{g}$

The standard expression for the horizontal range of a projectile on level ground is $R=\dfrac{v_0^2\sin2	heta}{g}$.
**Option A is the time of flight, and option C is related to maximum height.**

Answer

$R=\dfrac{2v_0\sin	heta}{g}$

Answer

$R=\dfrac{v_0^2\sin^2	heta}{2g}$

Answer

$R=\dfrac{v_0\cos	heta}{g}$

For a projectile launched with speed $v_0$ at angle $\theta$, the horizontal range is

Keep practising Unit 1A: Kinematics: 1D Motion, Position & Velocity