Question

a projectile is launched from and returns to ground level, as the figure shows. there is no air resistance. the horizontal range of the projectile is measured to be r = 174 m, and the horizontal component of the launch velocity is v0x = 29.0 m/s. find the vertical component v0y of the projectile.

Explanation:

Step1: Find the total time of flight

The horizontal motion is a uniform - motion with $R = v_{0x}t$. So, $t=\frac{R}{v_{0x}}$.
$t=\frac{174}{29.0}=6\ s$

Step2: Analyze the vertical motion

The time of flight $t$ for a projectile launched and landing at the same height is given by the formula $t = \frac{2v_{0y}}{g}$ (where $g = 9.8\ m/s^{2}$). We can re - arrange this formula to solve for $v_{0y}$.
$v_{0y}=\frac{gt}{2}$
Substitute $t = 6\ s$ and $g = 9.8\ m/s^{2}$ into the formula:
$v_{0y}=\frac{9.8\times6}{2}=29.4\ m/s$

Answer:

$29.4\ m/s$