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.

Answer 1

Time to fall: t = sqrt(2h/g) = sqrt(2(8.0)/9.8) = 1.2778 s (≈).
Horizontal speed: v = d/t = 6.5 / 1.2778 = 5.087 m/s → 5.09 m/s (to nearest hundredth).