### Video Transcript

A child who has a mass of 30 kilograms jumps into the air and lands on the ground. The child hits the ground at a speed of one metre per second and comes to rest 0.1 seconds after landing. What is the child’s weight what is the maximum force applied to the ground by the child’s feet during the landing?

Okay, so in this question, we’ve got a child who has a mass of 30 kilograms. Just before landing, the speed of the child is one metre per second. And it takes 0.1 seconds for the child to stop. We need to find out the child’s weight and also the maximum force applied to the ground by the child’s feet during the landing.

Let’s start by finding out the child’s weight. We need to find a relationship between the mass of the child and the weight of the child. This is the relationship we’re looking for: the weight 𝑊 is equal to the mass 𝑚 multiplied by the gravitational field strength of the Earth 𝑔.

Before we plug in any values there, let’s quickly discuss the standard units of this equation. The standard unit of mass is kilograms and the standard unit of the gravitational field strength of Earth is newtons per kilogram. Since we’re multiplying the two quantities together, these two will cancel leaving us with a total unit of newtons.

In other words, the standard unit of weight is newtons. And this is exactly what we’d expect because weight is a force. So if we provide this equation with the mass in kilograms and the gravitational field strength in newtons per kilogram, then we’ll get the weight out in newtons.

So let’s do that now. The weight of the child 𝑊 is equal to the mass of the child 30 kilograms multiplied by the gravitational field strength of the Earth which is 9.8 newtons per kilogram. Evaluating this gives us 294 newtons. Therefore, the child’s weight is 294 newtons.

Moving on, we need to find out the maximum force applied to the ground by the child’s feet during the landing. Here’s a diagram showing the child just about to land. We know that just before the landing, the child has a speed of one metre per second downwards. And the child is stationary 0.1 seconds later. So with this information, we can work out the deceleration of the child.

We’ll label this quantity 𝑎 because deceleration is the same thing as an acceleration in the opposite direction. Therefore, we can work out the acceleration of the child, which happens to be the following: an acceleration is defined as the change in velocity divided by the time taken for that velocity change to occur. In symbols, this is Δ𝑣 divided by 𝑡 because Δ represents a changing. Therefore, Δ𝑣 is a change in velocity and the time is represented by 𝑡.

First things first, the change in velocity, well, we know the final velocity of the child. This happens to be zero metres per second. And we also know the initial velocity of the child which is one metre per second. In order to find the change in velocity, we need to find the difference between the final velocity and the initial velocity. So that’s zero metres per second minus one metre per second.

And in order to find the acceleration, we need to divide this by the time taken for that change in velocity to occur. This time is 0.1 seconds. So we can evaluate this fraction, which gives us an acceleration for the child of negative 10 metres per second squared. Now, the negative sign simply represents the fact that the child is accelerating in the opposite direction to his initial motion. In other words, the child was travelling downwards and the acceleration is in the opposite direction. This makes sense. The child is slowing down after all.

So now, what can we do? Well, since we’ve got the acceleration of the child, we can work out the force exerted on the child to cause this acceleration. We can use Newton’s second law, which tells us that the force exerted on an object 𝐹 is equal to the mass of that object 𝑚 multiplied by its acceleration 𝑎.

And very quickly, let’s discuss the units. If we want the force exerted on an object to be in newtons, then we have to provide the equation with a mass in kilograms and an acceleration in metres per second squared. Luckily, our mass and acceleration are in the exact correct quantities. So the force exerted on the child is 30 kilograms multiplied by negative 10 metres per second squared. And as we’ve just said, this force is gonna be in newtons. So the force happens to be negative 300 newtons.

Once again, the negative value suggests to us that the force is acting in the opposite direction to the direction of motion. The child is falling downwards and the force decelerating it is acting upwards. However, we need to be very careful here. This force that we’ve calculated is the resultant force on the child. In other words, it’s the total force on the child when we’ve accounted for all of the individual forces acting on the child.

So let’s put the negative 300 newtons force here on the left, where we’ve put the acceleration as well. And let’s study the individual forces acting on the child. Now, here’s the child just about to land and there’s a ground. Now, we’ve worked out that the resultant, the total, the net force — whatever you want to call it — is 300 newtons in the upward direction. So let’s label the individual forces on the child.

One of them we’ve worked out already — it’s the child’s weight. The weight obviously acts in a downward direction and we calculated it earlier — that is 294 newtons. But because the child is landing on the ground, there must be another upward force. This is the force exerted by the ground on the child. We don’t know what this force is. So we’ll just call it 𝑥 newtons. Now, it’s the 𝑥 newton force in the upward direction combined with the 294-newton downward direction force. That gives us the resultant of 300 newtons.

So we can calculate the value of 𝑥 this way: negative 300 newtons — that’s the resultant force — is equal to negative 𝑥 plus 294. So it’s the 𝑥 in the upward direction we’ve taken it to be negative once again. And since the 294-newton force is in the downward direction, that’s positive. We can now rearrange the equation by adding 𝑥 to both sides and 300 to both sides as well. That leaves us with 𝑥 is equal to 300 plus 294. And so we get that the value of 𝑥 is 594.

Now, you might be wondering, why this value doesn’t end up being negative? Well, we’d already accounted for the fact that the 𝑥 newton force is negative up here. So we can simply replace the 𝑥 newtons up here with 594 newtons. Now, this is the force as we’ve said earlier that the ground exerts on the child. But if we read the question carefully, it wants us to calculate the force exerted by the child on the ground. So how do we go about calculating this?

Well, luckily, there’s that little thing known as Newton’s third law of motion. This states that if we got two objects — objects one and two — and object one exerts a force on object two, then object two exerts an equal and opposite force on object one. Applying this to our scenario, object one is the ground. That is exerting a force of 594 newtons upwards on the child, which is object two. Therefore, object two — the child — exerts an equal force, but in the opposite direction on object one.

In other words, the child exerts a 594-newton force downwards on the ground. And obviously, this force is gonna be exerted by the child’s feet because it is the feet of the child that make contact with the ground.

Now, the question says “what is the maximum force applied during the landing?” Well, here we’ve assumed that the force applied by the child on the ground is constant. So the maximum force is simply the force we’ve calculated. And so our final answer is that the force applied by the child’s feet to the ground during the landing is 594 newtons.