Physics

Problem

Which one of the following expressions gives the vertical displacement

y

of a projectile, launched from the origin with initial speed

v_0

at an angle

\theta_0

to the horizontal, in terms of the horizontal displacement

x

and acceleration due to gravity

g

? a)

y = x\tan\theta_0 - \dfrac{g x^2}{2 v_0^2 \sin^2\theta_0}

y = x\tan\theta_0 - \dfrac{g x^2}{v_0^2 \cos^2\theta_0}

y = x\tan\theta_0 - \dfrac{g x^2}{2 v_0^2 \cos^2\theta_0}

y = x\tan\theta_0 + \dfrac{g x^2}{2 v_0^2 \cos^2\theta_0}

Intuitive explanation

The key is to eliminate time from the kinematic equations by using the horizontal motion to express $t$ in terms of $x$ , then substitute into the vertical equation. Since horizontal velocity is constant ( $v_x = v_0 \cos\theta_0$ ), you can write $t = \frac{x}{v_0\cos\theta_0}$ . Then plug this into the vertical displacement equation $y = v_0\sin\theta_0 \cdot t - \frac{1}{2}gt^2$ to get $y$ as a function of $x$ alone—this gives you the trajectory equation.

Intuitive explanation

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