projectile is shot from cliff 125.0m above ground w/ Vo = 65.0m/s at 37.0° w/ the horiz. speed just before hitting ground? (no kinematics)?

Conservation of energy shall apply :

m*Vo^2+2*m*g*h = m*Vf^2

mass m cross

Vf = √Vo^2+2*g*h = √65*2+250*g = √65*2+250*g = 81.71 m/sec

...no matter the shooting angle...what matter are just initial speed module and height change

Since the projectile has a vertical and horizontal velocity, the only way that I know how to solve this problem is by using the kinematic equations. The first step is to determine the projectile’s initial vertical and horizontal velocity.

Vertical = 65 * sin 37 ≈ 39 m/s

Horizontal = 65 * cos 37 ≈ 52 m/s

During the time the projectile is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the vertical velocity as it hits the ground.

vf^2 = vi^2 + 2 * a * d, a = -9.8 m/s^2

d is the projectile’s vertical displacement.

d = final height – initial height = 0 – 125 = -125 meters.

vf^2 = (65 * sin 37)^2 + 2 * -9.8 * -125

vf = ± √(65^2 * sin^2 37 + 2,450).

This is approximately 63.1 m/s. Since the projectile is falling, its vertical velocity as it hits the ground is negative. The projectile’s horizontal velocity has not changed. Let’s use the following equation to determine the magnitude of the projectile’s velocity as it hits the ground.

Speed = √(Vertical^2 + Horizontal^2)

Speed = √[(65^2 * sin^2 37 + 2,450) + (65 * cos 37)^2 = 6675

This is approximately 81.7 m/s. If you need to determine the angle of the projectile’s velocity as it hits the ground, use the following equation.

Tan θ = Vertical ÷ Horizontal = -√(65^2 * sin^2 37 + 2,450) ÷ (65 * cos 37)

This is approximately -51˚. The negative sign means the projectile is moving downward and horizontally as it hits the ground. I hope this is helpful for you.

Vertical velocity is Vy = 65 sin 37 = 39.12 m/s

It will raise to a height of

h = v²/2g = 39.12²/2•9.8 = 78.07 m

with zero speed

then it falls 78.07 + 125 m down

velocity = Vy = √(2gh) = √(2•9.8(125+78.07)) = 63.09 m/s

Vx = 65 cos 37 = 51.91 m/s

which is the same as when it lands

V = √(Vx² + Vy²)

V = √(51.91² + 63.09²) = 81.7 m/s