Projectile Motion Notes

Projectile Motion - Free fall with an initial horizontal velocity.

Projectiles launched horizontally

Projectiles launched at an angle

Projectiles have two components

x = horizontal component

y = vertical component

 $V_{y,\ f}=a_y\Delta t$  : The final velocity in the y direction is equal to the acceleration in the y direction (gravity) times the change in time.

 $\left(V_{y,\ f}\right)^2=2a_y\Delta t$  : The final velocity in the y direction squared is equal to double the acceleration in the y direction (gravity) times the change in time.

 $\Delta y=\frac{1}{2}a_y\left(\Delta t\right)^2$  : Displacement in the y direction is equal to 1/2 the acceleration due to gravity times the change in time squared.

 $V_x=V_{x,\ i}$  : Horizontal velocity at any point is equal to the initial horizontal velocity. This means the velocity in the x direction is constant!

 $\Delta x=V_x\Delta t$  Displacement in the x direction is equal to the velocity in the x direction times the change in time.

Suppose you kick a rock off a 321 m tall bridge. The magnitude of the rock's horizontal displacement is 45 m.

Find the speed at which the rock was kicked.

Jenny rides her bike off an apartment building in an attempt to land in the pool. The four-story building is 12.5m tall, and she rides at a top speed of 11m/s. 

How long is she in the air?

If the pool is 5m from the building, and is 10m wide, will she splash down, land short, or fly right over the pool?

Stephanie goes cliff diving. The cliff is 52 meters high. How long would it take for her to hit the water?

There is a 6.5 meter wide sandbar directly below the cliff. What must her horizontal speed be to clear the sandbar and hit the deep water?

 $V_{y,\ f}=V_i\ \sin\theta\ +a_y\Delta t$  : Vertical final velocity is equal to the initial velocity times the sine of the launched angle plus the vertical acceleration times the change in time.

 $\left(V_{y,\ f}\right)^2=V_i^2\ \left(\sin\theta\right)^2+2a_y\Delta y$  : Final vertical velocity squared = initial velocity squared times the sine of the launch angle squared + 2 times the vertical acceleration times the vertical displacement.

 $\Delta y=\left(V_i\ \sin\theta\right)\Delta t+\frac{1}{2}a_y\left(\Delta t\right)^2$  : Vertical displacement = initial velocity times the sine of the launch angle times the change in time + 1/2 vertical acceleration times the change in time squared.

 $V_x=V_{x,\ i}=V_i\ \cos\theta\$  : The velocity in the horizontal direction is constant. 

The x-component for velocity can be found by multiplying the initial velocity by the cosine of the launch angle.