1. Missile Launch Formulas and examples

2. Velocity in x and y direction General: X component of original velocity: v ox = v o cos(theta) Y component of original velocity: v oy = v o sin(theta)

3. θ

4. Example 1 In the x direction: v ox = v o cos(theta) v ox = (40.0 m/s)(cos(35 degrees)) v ox = (40.0)(0.8191) v ox = 32.76 v ox = 32.8 m/s

5. Example 2 In the y direction: v oy = v o sin(theta) v oy = (40.0 m/s)(sin(35 degrees)) v oy = (40.0)(0.5735) v oy = 22.94 v oy = 22.9 m/s

6. Time at the top General: We can use the following kinematics equation: v f = v o + at Subscript it for y: v fy = v oy + a y t Solve it for t: t = (v fy - v oy ) / a y Plug in 0.0 m/s for v fy : t = (0.0 m/s - v oy ) / a y

7. Example Start with: t = (v fy - v oy ) / a y Plug in 0.0 m/s for v fy : t = (0.0 m/s - v oy ) / a y Plug in values for v oy and a y : t = (0.0 m/s - 22.9 m/s) / - 9.8 m/s 2 t = -22.9 / -9.8 t = 2.33 t = 2.3 s In this example 2.3s of time passes while the projectile is rising to the top of the trajectory.

8. Displacement General: Here is the displacement formula: d = v o t + 0.5at 2 We must think of this displacement in the y direction, so we will subscript this formula for y: d y = v oy t + 0.5a y t 2 If now we plug in the half time of flight, which was found above, we will solve for the height of the trajectory, since the projectile is at its maximum height at this time.

9. Example Starting with: d y = v oy t + 0.5a y t 2 Then plugging in known values: d y = (22.9 m/s)(2.33 s) + (0.5)(-9.8 m/s 2 )(2.33 s) 2 d y = 53.35 - 26.60 d y = 26.75 d y = 27 m