### Video Transcript

A frog jumps from the ground and lands at a point 16 centimeters horizontally distant from its launch point, moving with a constant horizontal velocity. The maximum vertical displacement of the frog above the ground during its leap is 3.6 centimeters. What is the horizontal velocity of the frog?

Letβs start with a diagram of this scenario. We have a frog that jumps from the ground. The frog will have an initial speed that we will call π£ at an angle above the horizontal that we will call π. We are told that the frog moves through the air with a constant horizontal velocity. So the only force acting on the frog while it is in the air must be from gravity. This force is the weight of the frog and has a magnitude of the mass of the frog, which we will call π, multiplied by the acceleration due to gravity, which is π.

Because the only force acting on the frog while it is in the air is gravity, it will behave as a projectile. This means that the frogβs path through the air will be curved. It will travel upwards, reaching a maximum vertical displacement, also known as its maximum altitude. And then it will travel downwards to the point where it lands, which we are told in the question is horizontally distant from its launch point. So the point where it lands has the same vertical displacement as its launch point. We will label the maximum altitude of the frog β and the horizontal distance from its launch point to its landing point capital π
. And this is also known as the projectileβs horizontal range.

The question tells us that the frogβs horizontal range is 16 centimeters and the maximum altitude of the frog is 3.6 centimeters. The question wants us to use these to find the horizontal velocity of the frog. We will start by looking at equations for the maximum altitude and horizontal range of a projectile. Letβs start with the equation for horizontal range, which states that horizontal range is equal to two multiplied by the initial speed of the projectile squared multiplied by the cosine of the launch angle multiplied by the sine of the launch angle divided by the acceleration due to gravity.

Next, letβs look at the equation for the maximum altitude of a projectile, which states that maximum altitude is equal to the initial speed of the projectile squared multiplied by the sine of the launch angle of the projectile squared divided by two multiplied by the acceleration due to gravity. But how do these two equations relate to the horizontal velocity of the frog?

To find out, letβs look at a diagram of the initial speed of the frog. Weβve stated that the frog jumps from the ground with an initial speed of π£ at an angle of π above the horizontal. We can see that this initial speed has a horizontal component that we will call π£ π₯ and a vertical component that we will call π£ π¦. These form a right-angled triangle, so we can write that the horizontal velocity of the frog, π£ π₯, is equal to the initial speed of the frog, π£, multiplied by the cosine of its launch angle above the horizontal, π. And the initial vertical velocity of the frog, π£ π¦, is equal to the initial speed of the frog, π£, multiplied by the sine of the launch angle above the horizontal, π.

Now these look similar to some of the terms that we can see in our equations for horizontal range and maximum altitude. Letβs see if we can write our equation for the horizontal range of the frog in a way that uses π£ multiplied by the cos of π and π£ multiplied by the sin of π, starting with our initial expression of π
equals two π£ squared multiplied by the cos of π multiplied by the sin of π divided by π. And notice that π£ squared can be written as π£ multiplied by π£. And we are free to change the order of multiplication, so we can bring one π£ to in front of the sin π term. So just by slightly rearranging our equation for horizontal range, we have it in terms of horizontal velocity and vertical velocity. And instead of π£ multiplied by the cos of π, we can write π£ π₯. And instead of π£ multiplied by the sin of π, we can write π£ π¦.

So we have an expression for the horizontal range of the frog in terms of the frogβs horizontal velocity and its initial vertical velocity and acceleration due to gravity. We already know the horizontal range of the frog, and we know acceleration due to gravity. So in order to work out the horizontal velocity of the frog, which is our goal, we must first work out the initial vertical velocity of the frog, π£ π¦.

Before we start trying to work out the initial vertical velocity of the frog, we will keep a note of this equation for horizontal range in terms of π£ π₯ and π£ π¦ in place of our original equation for horizontal range. To find an expression for the initial vertical velocity of the frog, we can look at our equation for its maximum altitude. Starting with our initial expression, β is equal to π£ squared multiplied by sin π squared divided by two π, we notice that π£ squared multiplied by sin π squared is equal to π£ multiplied by sin π all squared. So we can rewrite our expression for maximum altitude to β is equal to π£ multiplied by the sin of π all squared divided by two π. And we know that π£ multiplied by the sin of π is equal to the initial vertical velocity of the frog. So we can actually write this as β is equal to π£ π¦ squared divided by two π.

Weβre looking to find an expression for the initial vertical velocity of the frog. So we need to rearrange this equation for π£ π¦. We will start by multiplying both sides by two π, and both the twos and the πβs will cancel on the right-hand side. Next, we will take the square root of both sides. And the square root of π£ π¦ squared is just π£ π¦, giving us our expression for the initial vertical velocity of the frog, which we will keep a note of over here on the left. Looking at our equation for the horizontal range of the frog, we know the horizontal range. We also know the acceleration due to gravity, and we now have an expression for the initial vertical velocity of the frog, π£ π¦.

All thatβs left to do is rearrange this equation to get an expression for the horizontal velocity of the frog, the π₯. We will start by multiplying both sides of the equation by π. And these πβs on the right will cancel. We will then divide both sides of the equation by two, and the twos on the right will cancel. Finally, we will divide both sides of the equation by π£ π¦, where the π£ π¦βs on the right will cancel. So we have an expression for π£ π₯, the horizontal velocity of the frog.

Finally, letβs substitute our expression for the initial vertical velocity of the frog into this equation. And this is our final expression for the horizontal velocity of the frog. Letβs write this a bit more neatly. The horizontal velocity of the frog π£ π₯ is equal to the acceleration due to gravity, π, multiplied by the horizontal range of the frog capital, π
, divided by two multiplied by the square root of two multiplied by the acceleration due to gravity, which is π, multiplied by the maximum altitude of the frog, which is β.

We know that π
is equal to 16 centimeters, β is equal to 3.6 centimeters, and π is equal to 9.8 meters per second squared. Before we substitute any of these values in, we must check that they are all in SI units. This means that any lengths should be in units of meters instead of centimeters. So π
is equal to 0.16 meters and β is equal to 0.036 meters. Substituting these into our equation, we get π£ π₯ is equal to 9.8 meters per second squared multiplied by 0.16 meters divided by two multiplied by the square root of two multiplied by 9.8 meters per second squared multiplied by 0.036 meters. Evaluating this entire expression, we get π£ π₯ is equal to 0.93 meters per second to two decimal places. The horizontal velocity of the frog is 0.93 meters per second.