You are standing on the edge of a ravine that is 17.5 m wide. you notice a cave on the opposite wall whose ceiling is 7.3 m below your feet. the cave is 4.2 m deep, and has a vertical back wall. you decide to kick a rock across the ravine into the cave. what initial horizontal velocity must you give the rock so that the rock barely misses the overhang? the acceleration of gravity is 9.8 m/s 2 .

v₀ₓ = 14.34 m / s

Explanation:

We can solve this problem using the projectile launch equations.

Let's look for the time it takes to descend to the height of the cave

    y = t - ½ g t²

As it rises horizontally the initial vertical speed is zero  

    y = 0 - ½ gt²

    t = √2 y / g

    t = √2 7.3 / 9-8

    t = 1.22 s

This is the same time to cross the ravine

    x = v₀ₓ t

    v₀ₓ = x / t

    v₀ₓ = 17.5 / 1.22

    v₀ₓ = 14.34 m / s

This is the minimum speed.

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000