A projectile is launched with initial speed $v_0$ at angle $ heta$. I…

Question

A projectile is launched with initial speed $v_0$ at angle $	heta$. Its maximum height is given by

Accepted Answer

C. $H=\dfrac{v_0^2\sin^2	heta}{2g}$

The vertical component of launch speed is $v_0\sin	heta$.
Using vertical motion, the maximum height is $H=\dfrac{(v_0\sin	heta)^2}{2g}=\dfrac{v_0^2\sin^2	heta}{2g}$.

Answer

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

Answer

$H=\dfrac{v_0^2\cos^2	heta}{2g}$

Answer

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

A projectile is launched with initial speed $v_0$ at angle $\theta$. Its maximum height is given by

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