Video: 2D Projectile Motion

An archer shot an arrow with a speed of 80 m/s at an angle of 20°‎ above the horizontal. Calculate the arrow’s horizontal range 𝑥. Also calculate its vertical height ℎ above the launch point when it was at a distance of 46 m from the archer. Give your answers in metres correct to one decimal place. Consider the acceleration due to gravity to be 9.8 m/s².

06:55

Video Transcript

An archer shot an arrow with a speed of 80 metres per second at an angle of 20 degrees above the horizontal. Calculate the arrow’s horizontal range 𝑥. Also calculate its vertical height ℎ above the launch point when it was at a distance of 46 metres from the archer. Give your answers in metres correct to one decimal place. Consider the acceleration due to gravity to be 9.8 metres per second squared.

When dealing with the projection of a particle, in this case an arrow, at an angle, we always look to break the motion into two parts. We want to consider the horizontal projection and the vertical projection of the arrow. And to do this, we can sketch the initial motion of the arrow out.

We’re going to add a right-angled triangle to the diagram. And we can see that this right-angled triangle has a hypotenuse of 80 metres per second and an included angle of 20 degrees. We’re looking to begin by modelling the horizontal and vertical components of the initial velocity of the arrow. And we can use right angle trigonometry to do so.

Remember the side directly opposite the included angle is the opposite side. And the side next to the included angle is the adjacent. We already said that the hypotenuse of this triangle was 80 metres per second. Let’s call the initial horizontal speed 𝑢𝑥 and the initial vertical speed 𝑢𝑦.

And since sin 𝜃 is equal to opposite over hypotenuse, we can say that sin of 20 degrees is equal to 𝑢𝑦 over 80. And if we multiply both sides of this equation by 80, we can see that 𝑢𝑦 is equal to 80 sin of 20 degrees. So, that’s the initial vertical speed. We know that cos of 𝜃 is equal to adjacent over hypotenuse, so we can say that cos of 20 degrees is equal to 𝑢𝑥 over 80. And once again, we multiply through by 80. And we can see that 80 cos of 20 degrees is equal to 𝑢𝑥. And we found the initial horizontal speed of the arrow.

Now let’s have a look at our question and see what we’re trying to find. We want to begin by finding the horizontal range of the arrow. Now if we assume that the arrow is shot from a point zero metres above the horizontal as in our diagram, we can say that the horizontal range will be found when the arrow reaches the horizontal again. In other words, when it hits for floor. That will be when the vertical height or displacement is zero. So, what else do we know?

We know that the acceleration due to gravity is 9.8 metres per second squared. We’ve assumed the arrow to be travelling upwards. So, the vertical acceleration is negative 9.8. It’s acting against the direction that the arrow’s currently travelling. So, we have the initial vertical speed, the acceleration, and the height we’re aiming for. In order to find a horizontal range, we’re going to need to find the time taken for it to reach this height.

So, we recall the equations of motion. They are 𝑣 equals 𝑢 plus 𝑎𝑡, 𝑠 equals 𝑢𝑡 plus half 𝑎𝑡 squared, 𝑠 equals a half multiplied by 𝑢 plus 𝑣 multiplied by 𝑡, 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠, and 𝑠 equals 𝑣𝑡 minus a half 𝑎𝑡 squared. We’re going to use the one which does not have a final velocity 𝑣. That’s this one. 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared.

We can substitute what we know about the vertical motion of the arrow into this formula. We get zero equals 80 sin of 20 degrees multiplied by 𝑡 plus a half multiplied by negative 9.8 multiplied by 𝑡 squared. A half of negative 9.8 is negative 4.9. And we can solve this by factoring the expression in 𝑡. And it doesn’t matter if we don’t actually fully factor, instead we’ll just take out a factor of 𝑡.

We get zero is equal to 𝑡 multiplied by 80 sin of 20 degrees minus 4.9𝑡. 𝑡 is multiplying everything inside these brackets. And when it does, it gets an answer of zero. That means that either 𝑡 must be equal to zero, or the expression inside the brackets must be equal to zero. Now we already know that the starting height of the arrow was zero, so we’re going to disregard 𝑡 equals zero. Let’s solve the second equation.

We begin by adding 4.9𝑡 to both sides and then we divide through by 4.9. And we get 𝑡 is equal to 5.58 and so on. So, we now need to find the horizontal displacement when 𝑡 is equal to 5.58. And we’ll use the unrounded version. We know that the horizontal component for the velocity has no acceleration. It has a constant velocity. So, 𝑣 in the horizontal direction is always 80 cos 20. We know that 𝑡 is equal to 5.58 seconds. And we’re looking to find the horizontal range. We’re looking to find the horizontal displacement 𝑠.

We can actually use a whole number of equations here, but let’s use 𝑠 equals a half 𝑢 plus 𝑣 multiplied by 𝑡. We get a half multiplied by 80 cos of 20 degrees plus 80 cos of 20 degrees multiplied by 5.58 and so on. And if we type that into our calculator, we get 419.779 and so on. Correct to one decimal place, the horizontal range of the arrow is 419.8 metres.

Next, we want to calculate the vertical height ℎ above the launch point when it’s at a distance of 46 metres from the archer. This time the horizontal displacement 𝑠 is 46. We don’t know the time it takes to reach this point. And we’re looking to find the horizontal height 𝑠 equals ℎ. This time we’ll substitute what we know into the third formula, 𝑠 equals a half 𝑢 plus 𝑣𝑡. We’re trying to find the time it takes to reach a horizontal distance of 46 metres from the archer.

In fact, we had two lots of 80 cos of 20 and then we wanted half of that, which is simply 80 cos of 20. So, we can find the value of 𝑡 by dividing 46 by 80 cos of 20. This time that gives us a value of 0.6119 and so on seconds. We now know the initial vertical speed, we’re trying to calculate the height. We know the acceleration and the time.

This time we’ll use a formula 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. That’s ℎ equals 80 sin of 20 degrees multiplied by 0.6119 plus a half multiplied by negative 9.8 multiplied by, once again, that unrounded number 0.6119. That gives us 14.907 and so on. This time, correct to one decimal place, that’s 14.9 metres. The horizontal range is 419.8 metres. So, in fact, we can say 𝑥 equals 419.8 metres and ℎ equals 14.9 metres.

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