how to find speed with vertical height

Question

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v₀ = 16 m/s at an angle θ = 32 ° below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone's initial position.

With what speed, vf in meters per second, does the stone strike the ground?

Answer 1

Step 1: Find the x and y components of the initial velocity upon throwing the stone off of the cliff.

v_0,x = v₀cos(θ) = 16cos(32) = 13.57m/s

v_0,y = v₀sin(θ) = 16sin(32) = -8.48m/s (I'm considering down the negative y direction)

Step 2: Since there are no forces acting in the x direction, the initial velocity in the x direction will be as the final velocity in the x direction. The next piece of information we need is the final velocity in the y direction. Let's use the following kinematic equation.

v_f,y = v_0,y + a_yt

v_f,y = Final velocity of object in y-direction

v_0,y = Initial velocity of object in y-direction = -8.48m/s

a_y = Acceleration in the y-direction = -9.81m/s²

t = time

This looks like the right direction to find final velocity in the y-direction, but we also need time to solve the equation. Let's use another kinematic expression to find the time it took for the stone to fall to the ground.

x_f,y = x_0,y + v_0,yt + 0.5a_yt²

x_f,y = Final position of object in y-direction = 0m

x_0,y = Initial position of object in y-direction = 8m

v_0,y = Initial velocity of object in y-direction = -8.48m/s

a_y = Acceleration in the y-direction = -9.81m/s²

t = time

Use quadratic formula to solve for time.

t = -2.41s, 0.68s (Drop the negative term since time can't be negative)

t = 0.68s

Plug this back into the other equation to find the final velocity in the y-direction.

v_f,y = -15.15m/s

Step 3: Use the components of the final velocity to calculate the magnitude of the final velocity.

v_f = √[(v_f,y²) + (v_f,x²)] = √((-15.15)² + 13.57²) = 20.34m/s