# How much force would Superman need to exert on the ground in order to just reach the top of the Burj Khalifa building (while jumping straight upwards). Assume that his feet stay in contact with...

How much force would Superman need to exert on the ground in order to just reach the top of the Burj Khalifa building (while jumping straight upwards). Assume that his feet stay in contact with the ground for 1 second during his jump that his mass is 100 kg and that air resistance is negligible.

Superman has the reputation of being able to "leap tall building is a single bound". Suppose Superman wants to prove he can do this for the Burj Khalifa tower, the most recent record holder for the worlds tallest building at a height of 828 meters.

I need help with this practice problem that I just cant seem to get. I tried using conservation of energy, but it requires a force. Any help would be much appreciated!

The first thing you need to do is set up your equation. You are looking for the force which is needed to jump vertically (overcoming the force of gravity pointing in the opposite direction) a height of 828 meters.

The first equation we need to define is Newtons Second Law which states that when a net force is applied to a body of mass, the body will be accelerated in the same direction as the force applied to the mass which is expressed as Force (F) = mass (m) x acceleration (a).

We also need to be aware of the formula which describes acceleration which is expressed as: acceleration (a) = meters (m) / second (s) ^2.

Finally we need to account for the gravitational force Superman must overcome in order to "jump" off the ground: Fg = (m)(a), the gravitational acceleration of earth is a constant = 9.8 m/s^2.

Now we need to combine all the equations to solve for the force required => F= (mass x meter/second^2) - (mass * gravitational acceleration)

This gives us F = (100kg * 828m / 1^2) - (100kg * 9.8m/s^2) = 81,820 kg*m /s^2 or 81,820 Newtons (N) of force which is required for Superman to jump the Burj Khalifa Building.

In this case, we consider that the impulse of the force (FΔt), is equal to the variation in the amount of body movement (mΔv).

F(t –t0) = m(v – v0) (1)

Where:

t –t0 → is the time that is in contact with the ground.

v → is the final speed at the moment it leaves the ground.

v0 → is the initial velocity, before applying the force on the ground.

Before applying the force on the ground, the velocity v0 is zero. To find the final velocity v, when it leaves the ground, we can apply the law of conservation of energy; that is, the kinetic energy (mv0^2/2), when it leaves the ground, is equal to the gravitational potential energy (mgh) at the top:

mv^2/2 = mgh → v = √(2gh)

v = √(2(9.8)828) = 127.4 m/s

Now we get the force, from equation (1):

F = m(v – v0)/(t –t0)

F = (100 kg)(127.4 m/s)/(1 s)

F = 12740 N

**Superman must apply a force of 12740 N on the ground, to reach the top of the building**.

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