Horizontal force applied to push box off 12.96 m cliff

[paigegail]

HELP! Projectile motion and Force puzzler!

A 10.8 kg box is 5.0 m from the edge of a 12.96 m high cliff. Bob pushes the box over the edge of the cliff by applying a constant horizontal force to it until the box goes over the edge. The force of friction on the box is 49.0 N and the box lands 7.5 m from the base of the cliff. How much force did Bob apply to the box while he was pushing?

What am i supposed to do... I know I'm supposed to work backwards...which is what I did. I found the velocities in the vertical and horizontal directions...what do I do now?!

 

[gnome]

If you know the velocity (horizontal) of the box right at the edge of the cliff, the next step is to solve for the NET force that had to be applied constantly over a distance of 5 m to accelerate it from rest up to that velocity.

You are given the force of friction, which was a constant force acting in the opposite direction, so how much force did Bob apply?

 

[M98Ranger]

First, let's break down the problem into smaller parts to make it more manageable. We know that the box was pushed horizontally until it went over the edge of the cliff, so let's focus on the horizontal motion first.

The box is initially at rest, so its initial horizontal velocity is 0 m/s. We also know that the box lands 7.5 m from the base of the cliff, so its final horizontal displacement is 7.5 m. We can use the equation for horizontal displacement to find the average horizontal velocity of the box:

d = (v0 + vf)/2 * t

Where:
d = displacement (7.5 m)
v0 = initial velocity (0 m/s)
vf = final velocity (unknown)
t = time (unknown)

Rearranging the equation, we get:

vf = (2d)/t

Now, let's focus on the vertical motion. We know that the box falls from a height of 12.96 m, so we can use the equation for vertical displacement to find the time it takes for the box to fall:

d = v0 * t + (1/2) * a * t^2

Where:
d = displacement (12.96 m)
v0 = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)

Rearranging the equation, we get:

t = √(2d/a)

Plugging in the values, we get:

t = √(2*12.96/-9.8) = 1.8 seconds

Now, we can use the equation for final velocity to find the vertical velocity of the box when it lands:

vf = v0 + a * t

Where:
vf = final velocity (unknown)
v0 = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (1.8 seconds)

Plugging in the values, we get:

vf = v0 + (-9.8 * 1.8) = v0 - 17.64

We also know that the vertical velocity when the box lands is 0 m/s, so we can set vf to 0 and solve for v0:

0 = v0 - 17.64
v0 = 17.64 m/s

Now that we have the horizontal and vertical velocities, we