A projectile is launched with initial speed $v_0$ at angle $\theta$. Its maximum height is given by
A$H=\dfrac{v_0^2\sin\theta}{2g}$
B$H=\dfrac{v_0^2\cos^2\theta}{2g}$
C$H=\dfrac{v_0^2\sin^2\theta}{2g}$
D$H=\dfrac{2v_0\sin\theta}{g}$
Answer & Solution
Correct answer: C. $H=\dfrac{v_0^2\sin^2\theta}{2g}$
The vertical component of launch speed is $v_0\sin\theta$. Using vertical motion, the maximum height is $H=\dfrac{(v_0\sin\theta)^2}{2g}=\dfrac{v_0^2\sin^2\theta}{2g}$.
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