You are standing at the top of a cliff that has a stairstep configuration. There

Question

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 4 m at your feet, then a horizontal shelf of 6 m, then another drop of 6 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.31 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.
initial horizontal velocity is 6.67 m/s
The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligi- ble.

How far from the bottom of the second cliff will the projectile land?
Answer in units of m

Explanation / Answer

to cross the step when the rock goes 4m down it should go at least 6m time = t so horizontally 6.67 * t = 6 => t=0.9 sec let it takes T time more to reach the bottom of 2nd cliff so (1/2) g(0.9 +T)^2 = 10 0.9 +T= 1.4286 T= .5286 sec so the distance from the bottom of the second cliff where the projectile will land is.5286*6.67 = 3.53m

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