1) A movie stunt driver speeds horizontally off a 50m high cliff. How fast must the motorcycle leave the cliff top if it is to land on the ground below, 90m from the base of the cliff?
2) A mail carrier leaves the post office and drives 22.0 km North. She then drives in a direction 30 degrees East of South for 47 km. What is her displacement from the post office?
Im thinking the pythagorean theorem has to do with both problems, but I'm not sure on how to apply it in each situation.
2) A mail carrier leaves the post office and drives 22.0 km North. She then drives in a direction 30 degrees East of South for 47 km. What is her displacement from the post office?
Im thinking the pythagorean theorem has to do with both problems, but I'm not sure on how to apply it in each situation.
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In projectile problems you separate the horizontal and vertical motions. The horizontal motion is constant velocity. The vertical motion is accelerated.
1. The vertical motion obeys d = (1/2)at^2 because it is accelerating at a constant rate and starting at 0 velocity (there's no vertical component initially).
Plug in the vertical distance d = 50m and solve for t. The acceleration is 9.8 m/sec^2.
The horizontal motion obeys d = v * t. Plug in the desired horizontal distance and the time you found out to fall 50 m, and solve for v.
2. That one is asking about the magnitude of a vector given two perpendicular components. That does indeed use the pythagorean theorem.
1. The vertical motion obeys d = (1/2)at^2 because it is accelerating at a constant rate and starting at 0 velocity (there's no vertical component initially).
Plug in the vertical distance d = 50m and solve for t. The acceleration is 9.8 m/sec^2.
The horizontal motion obeys d = v * t. Plug in the desired horizontal distance and the time you found out to fall 50 m, and solve for v.
2. That one is asking about the magnitude of a vector given two perpendicular components. That does indeed use the pythagorean theorem.