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 10 m, then another drop of 4 m to the bottom of the canyon, which has a horizontal floor. You kick a 0.76 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.

2.)How far from the bottom of the second cliff will the projectile land?

just crank 90's in fornite you will figure it out

To answer these questions, we need to analyze the motion of the projectile in two separate parts: the vertical motion and the horizontal motion.

1.) Let's start by finding the initial horizontal velocity required to barely clear the edge of the shelf.

Since there is no horizontal acceleration acting on the projectile, the horizontal velocity remains constant throughout its motion. The horizontal displacement of the projectile is equal to the distance from the top of the cliff to the edge of the shelf, which is 10 m.

We can use the kinematic equation for horizontal motion:
Δx = v_x * t
where Δx is the horizontal displacement, v_x is the horizontal velocity, and t is the time of flight.

Since the vertical motion only affects the time of flight (vertical motion does not influence horizontal velocity), we can use the time of flight obtained from the vertical motion analysis.

To analyze the vertical motion, we can use the equation relating vertical distance, initial velocity, time, and acceleration:
Δy = v_y * t + (1/2) * g * t^2
where Δy is the vertical displacement, v_y is the vertical velocity, g is the acceleration due to gravity, and t is the time of flight.

In this case, the vertical displacement is the total vertical drop, which is 6 m + 4 m = 10 m. The initial vertical velocity is 0 since the rock is only given an initial horizontal velocity. We can rearrange the equation to solve for t:
10 m = (1/2) * 9.8 m/s^2 * t^2
t^2 = (10 m) / (1/2 * 9.8 m/s^2)
t^2 = 2.04 s^2
t ≈ 1.43 s

Now we have the time of flight and can substitute it back into the horizontal motion equation to solve for the horizontal velocity:
10 m = v_x * 1.43 s
v_x = 10 m / 1.43 s
v_x ≈ 6.99 m/s

Therefore, the initial horizontal velocity required to barely clear the edge of the shelf is approximately 6.99 m/s.

2.) To find the horizontal distance from the bottom of the second cliff where the projectile will land, we can use the horizontal velocity and time of flight.

The horizontal distance traveled by the projectile is given by:
Δx = v_x * t

Substituting the values we found earlier:
Δx = 6.99 m/s * 1.43 s
Δx ≈ 10 m

Therefore, the projectile will land approximately 10 meters from the bottom of the second cliff.

@Steve, thank you so much!

it takes 1.107 s to fall 6m.

So, the horizontal velocity must be at least 10m/1.107s = 9.037 m/s

I think you can now do the rest. Figure how long it takes to fall the remaining 4 meters to the bottom of the canyon. Then use that to find the horizontal distance moved.