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

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

Centrapedial force is the answer

The boat's velocity relative to the riverbank can be resolved into two components. the component going north is 2.7 m/s and the component going west is 1.2 m/s.we add these two vectors and then find the magnitude of the resultant to get the speed.the magnitude of the resultant vector from adding can be obtained by the pythagorean theorem.    to two significant figures, the boat's speed relative to the riverbank is 3.0 m/s.