Non-Symmetrical Projectile Motion

1. Non-Symmetrical Projectile Motion • Conceptual Two Ways to Throw a Stone • 1.) From the top of a cliff, a person throws two stones. The stones • have identical initial speeds, but stone 1 is thrown downward • at some angle below the horizontal and stone 2 is thrown up at • the same angle above the horizontal. Neglecting air resistance, • which stone, if either, strikes the water with greater velocity? • Follow the general rules for projectile motion • Break the y-direction into parts • up and down • symmetrical only back to initial height. • Path 2: Find the answers for this stone we’re throwing off the roof. • Path 1: Find the answers for the same stone thrown as in path 1. Path 2. If you have trouble with the concept, then work the problem with numbers like those shown in this diagram. Path 1. Change this drawing to show what path 1 would look like, i.e., draw path 1 with numbers onto the diagram.

2. 2. An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. • a. At what horizontal distance from the skiers should the supplies be dropped? • b. Find the magnitude of the velocity of the box as it reaches the ground. • 3. A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff? • 4. An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find: • Its position and velocity after 8 s • b. The time required to reach its maximum height • c. The horizontal distance (range) • 5. A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s. • At what angle with was it released? • b. What was the maximum height achieved by the ball?

3. 2.An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. a. At what horizontal distance from the skiers should the supplies be dropped? vx = 80 m/s y = 250 m a= -9.80 m/s2 = 7.14 s x = vx t = (80)(7.14) = 571 m b. Find the magnitude of the velocity of the box as it reaches the ground. vx = 80 m/s vy = gt = (9.8)(7.14) = 70 m/s = 106 m/s

4. 3. A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff? vx = 15 m/s x = 47 m vy = 0 = 3.13 s y = ½ gt2 = ½ (9.8)(3.13)2 = 48 m

5. 4. An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find: a. Its position and velocity after 8 s vox = 100 cos 30 = 86.6 m/s voy = 100 sin 30 = 50 m/s vo = 100 m/s, 30 t = 8 s g = - 9.8 m/s2 x = vox t = 86.6(8) = 692.8 m y = voy t + ½ gt2 = 50(8) + ½ (-9.8)(8)2 = 86.4 m vx = vox = 86.6 m/s vy = voy + gt = 50 + (-9.8)(8) = - 28.4 m/s V = (86.62 + -28.42)1/2 V = 91.1 m/s θ= Inv Tan (-28.4/86.6) θ= -18.2

6. b. The time required to reach its maximum height At the top of the path: vy = 0 vy = voy + gt = 5.1 s c. The horizontal distance (range) Total time T = 2t = 2(5.1) = 10.2 s x = vox t = 86.6(10.2) = 883.7 m

7. 5. A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s. a. At what angle with respect to the horizontal was it released? vo = 15 m/s t = 2 s time to maximum height = 1 s at the top vy = 0 vy = voy + gt =40.8º

8. b. What was the maximum height achieved by the ball? y = voy t +½gt2 = (15)(sin40.8º)(1) + ½ (-9.8)(1)2 = 4.9 m