Projectile Motion

Question 11. Since the vertical component is zero at the top of the parabolic path, and the horizontal component is the same throughout, the minimum speed occurs at the point of maximum height.

Problems:

20. Cliff is 44.1 meters tall, lands 4.8 meters from the base of the cliff.

22.
The pebbles have a constant horizontal velocity, and zero vertical velocity when they hit the window.

Since the vertical displacement is 8.0 meters, the initial vertical speed is 12.52 m/s. This means (from v = v0 + a*delta_t) that the time to hit the window is 1.277 seconds.

Since delta_x = 9.0 meters, the horizontal component of velocity = 9.0 m / 1.277 seconds = 7.05 m/s. This is the speed of the pebbles hitting the window.

24. The drop takes 3.38 seconds using just the vertical direction. The horizontal speed (which stays constant) is 45 m / 3.38 seconds = 13.31 m/s.

26. It takes 1.228 seconds for the ball to reach its maximum, so it takes double this, or 2.456 seconds to fall back down to the ground.

27. The drop takes 1.621 seconds (since delta_x and vx are given.) Voy = 0 since it is thrown horizontally. The height delta_y of the building is then 12.88 meters.

28. You can set up an equation using the equation for delta_y and time.

-2.2 m = (14 m/s)sin(40)*t - .5*(9.8 m/s^2)*t^2

There are two solutions (times) when this is true, but only the positive root matters: this is at 2.055 seconds. The horizontal displacement is given by (14 m/s)*cos(40)*2.055 seconds = 22.04 meters.

Problem 5 from the handout:
vo = D/(2H/g)^1/2

WP3. Both bullets hit the ground at the same time.