a projectile is launched horizontally from a height of 8.0 m. the projectile travels 6.5 m before hitting…

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.

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.

Answer

Explanation:

Step1: Find time of flight

The vertical - motion of the projectile is a free - fall motion. The initial vertical velocity (v_{0y}=0\ m/s), the acceleration (a = 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 (v_{0y} = 0\ m/s), we get (y=\frac{1}{2}at^{2}). Then (t=\sqrt{\frac{-2y}{a}}). [t=\sqrt{\frac{-2\times(-8.0)}{9.8}}=\sqrt{\frac{16}{9.8}}\approx1.28\ s]

Step2: Find initial horizontal velocity

The horizontal motion of the projectile is a uniform - motion with constant velocity since there is no horizontal acceleration ((a_x = 0)). The horizontal displacement (x = 6.5\ m). Using the equation (x = v_{0x}t), we can solve for (v_{0x}). Then (v_{0x}=\frac{x}{t}). [v_{0x}=\frac{6.5}{1.28}\approx5.08\ m/s]

Answer:

5.08