The Trajectory of a Projectile (HSC SSCE Physics): Quizzes

Parabolic

Only $u_y$ and $y_0$

When $v_y = 0$

Remains constant

$45°$

$t = -\frac{2u_y}{g}$

$x_{\text{max}} = u_x t_{\text{flight}}$

$h = y_0 - \frac{u_y^2}{2g}$

Same as initial velocity

They achieve the same range