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you are standing at the top of a cliff that has a stairstep configuration. there is a vertical drop of 6 m at your feet, then a horizontal shelf of 6 m , then another drop of 4 m to the bottom of the canyon, which has a horizontal floor. you kick a 0.16 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.1. What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? The acceleration of gravity is 9.8 m/s2 . Consider air friction to be negligible. Answer in units of m/s.
2. How far from the bottom of the second cliff will the projectile land? Answer in units of m.



Answer :

1. The initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2.  The projectile will land at 9.54 m of the second cliff d = 9.54 m

1. To find the horizontal velocity of the rock we need to use the following equation:

d = v * t

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = √2d/g

g: is gravity = 9.8 m/s²

t = √2d/g  = 1.20s

Now, the horizontal velocity of the rock is:

v = d/t = 6.67m/s

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

To calculate the distance at which the projectile will land, first, we need to find the time:

t = √2d/g = 1.43s

the distance is:

d = v * t = 9.54m

Therefore, the projectile will land at 9.54 m of the second cliff.

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