Question

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

Answer 1

5.09 m/s

Step-by-step explanation:

The velocity of the projectile the moment it was launched, rounded to the nearest hundredth, is

5.09 m/s.

Correct On EDG

Answer 2

You can find

1) time to hit the ground
2) initial velocity
3) speed when it hits the ground

Equations

Vx = Vxo

x = Vx * t

Vy = Vyo + gt

Vyo = 0

Vy = gt

y = yo - Vyo - gt^2 / 2

=> yo - y = gt^2 / 2

1) time to hit the ground

=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2

=> t = √1.631 s^2 = 1.28 s

2) initial velocity

Vxo = x / t = 6.5m / 1.28s = 5.08 m/s

3) speed when it hits the ground

Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s

V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2

=> V = √ (183.56 m^2 / s^2) = 13.55 m/s