Question

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a horizontal shelf of 8 m, then another drop of 3 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.87 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

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

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

$d = v*t \rightarrow v = \frac{d}{t}$    

Where:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

Where:

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$      

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

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

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$    

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

I hope it helps you!