What initial horizontal velocity v will be required to barely clear the edge of the shelf below you? How far from the bottom of the second cliff will the projectile land?

Question

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 3 m at your feet, then a horizontal shelf of 6 m , then another drop of 7 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.74 kg rock, giving it an initial horizontal velocity that barely clears the shelf below.

Answer 1

Problem Overview

You’re standing on a cliff with a stairstep shape:

1. First vertical drop: 3 meters.
2. Horizontal shelf: 6 meters long.
3. Second vertical drop: 7 meters to the canyon floor.

A 0.74 kg rock is kicked horizontally from the top edge of the cliff and barely clears the edge of the shelf (6 m away, 3 m down).

Part 1: What initial horizontal velocity vxv_xvx​ is required to just clear the shelf?

The rock must:

1. Fall 3 m vertically (first drop),
2. Travel 6 m horizontally (length of the shelf),
3. In the same time.

We can find the time to fall 3 meters using the vertical motion equation:

y=12gt2⇒t=2yg=2×39.8≈0.612≈0.782 sy = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 3}{9.8}} \approx \sqrt{0.612} \approx 0.782\, \text{s}y=21​gt2⇒t=g2y​​=9.82×3​​≈0.612​≈0.782sNow, to find the horizontal velocity needed to travel 6 meters in that time:

x=vxt⇒vx=xt=60.782≈7.67 m/sx = v_x t \Rightarrow v_x = \frac{x}{t} = \frac{6}{0.782} \approx 7.67\, \text{m/s}x=vx​t⇒vx​=tx​=0.7826​≈7.67m/s✅ Answer (Part 1):

Initial horizontal velocity (v_x \approx 7.67, \text{m/s} to just clear the shelf.

Part 2: How far from the base of the second cliff does the rock land?

Now the rock falls another 7 m vertically from the shelf edge. Since the rock keeps the same horizontal velocity (7.67 m/s), we need to find:

1. Time to fall 7 m:
2. t=2yg=2×79.8≈1.429≈1.195 st = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 7}{9.8}} \approx \sqrt{1.429} \approx 1.195\, \text{s}t=g2y​​=9.82×7​​≈1.429​≈1.195sHorizontal distance traveled in this time:

x=vx×t=7.67×1.195≈9.17 mx = v_x \times t = 7.67 \times 1.195 \approx 9.17\, \text{m}x=vx​×t=7.67×1.195≈9.17m✅ Answer (Part 2):

The rock lands approximately 9.17 meters away from the base of the second cliff.