Question

a projectile is launched horizontally from a height of 8.0 m. the projectile travels 6.5 m before hitting the ground. the velocity of the projectile the moment it was launched, rounded to the nearest hundredth, is m/s.

Explanation:

Step1: Calculate time of fall

The vertical - motion of the projectile is a free - fall motion. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration due to gravity $g = 9.8\ m/s^{2}$, and the vertical displacement $y=- 8.0\ m$ (taking downwards as negative). Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, substituting the values we get $y=\frac{1}{2}gt^{2}$ (since $v_{0y} = 0$). Then $t=\sqrt{\frac{-2y}{g}}$.
$t=\sqrt{\frac{-2\times(-8.0)}{9.8}}=\sqrt{\frac{16}{9.8}}\approx1.28\ s$.

Step2: Calculate initial horizontal velocity

In horizontal motion (where there is no acceleration, $a_x = 0$), the horizontal displacement $x = 6.5\ m$. The horizontal velocity $v_x$ is constant, and we use the equation $x = v_x t$. Then $v_x=\frac{x}{t}$.
Substituting $x = 6.5\ m$ and $t\approx1.28\ s$, we get $v_x=\frac{6.5}{1.28}\approx5.08\ m/s$.

Answer:

$5.08$