Question Video: Calculating the Time of Flight of a Projectile Physics • First Year of Secondary School

Video Transcript

A projectile has an initial speed of 25 meters per second and is fired at an angle of 48 degrees above the horizontal. What is the time between the projectile leaving the ground and returning to the ground at the same height that it was launched from?

Let’s draw a diagram of this scenario. We have a projectile launched with an initial speed that we will call 𝑉, and it’s launched at an angle above the horizontal that we will call 𝜃. During projectile motion, there is only the downward force from gravity acting on the projectile, which has a magnitude of the mass of the projectile, which we will call 𝑚, multiplied by 𝑔, the acceleration due to gravity. This downward force causes the projectile’s path to be curved. And we are told that the projectile returns to the ground at the same height that it was launched from. We will call the vertical displacement of the projectile 𝑆 𝑦. And at the end of the motion, its vertical displacement is equal to zero.

The question asks us to work out the time at which it returns to the ground again, which we will call capital 𝑇, and is also known as the time of flight of the projectile. To answer this question, we need to find the time at which the projectile reaches a certain displacement. So we need an equation that relates displacement to the time during the projectile motion. We will start with the following formula, which tells us that the displacement of an object undergoing constant acceleration is equal to its initial speed multiplied by time plus one-half of the acceleration it is experiencing multiplied by time squared.

In our case, we want to look at the vertical displacement of the projectile, which we will call 𝑆 𝑦. So we can write 𝑆 𝑦 is equal to the projectile’s initial vertical velocity 𝑉 𝑦 multiplied by time minus one-half of 𝑔 multiplied by time squared. Notice that there is a negative sign here because gravity acts downwards as highlighted on our diagram. The point in the projectile’s motion that we’re interested in is when the vertical displacement of the projectile is equal to zero, so we can write zero equals 𝑉 𝑦 multiplied by 𝑡 minus a half 𝑔𝑡 squared. Then we can take 𝑡 as a common factor, giving zero equals 𝑉 𝑦 minus a half 𝑔𝑡 all multiplied by 𝑡.

This expression will equal zero at two values of 𝑡: one when 𝑡 equals zero and the other when the bit inside the brackets, 𝑉 𝑦 minus a half 𝑔𝑡, is equal to zero. This makes sense. At the beginning of the motion when 𝑡 equals zero, the vertical displacement of the projectile is also zero. The second time the vertical displacement of the projectile is zero is at the end of the motion. So we want to solve the following equation: zero equals 𝑉 𝑦 minus a half 𝑔𝑡. We’ll first add a half 𝑔𝑡 to both sides, and these terms on the right cancel each other out. Then we’ll multiply both sides by two, where these twos on the left cancel each other out. Finally, we’ll divide both sides of the equation by 𝑔, and these 𝑔’s on the left will cancel each other out.

Tidying this up slightly, we get 𝑡 equals two 𝑉 𝑦 divided by 𝑔, and this is equal to our time of flight capital 𝑇. All that’s left for us to do is to find an expression for the initial vertical velocity of the projectile 𝑉 𝑦. To do this, we can look at a diagram of the initial speed of the projectile and the angle that it was launched above the horizontal. The velocity of the projectile has a horizontal, which we will call 𝑉 𝑥, and a vertical, which we have called 𝑉 𝑦, component. 𝑉, 𝑉 𝑥, and 𝑉 𝑦 form a right-angled triangle. So we can see that 𝑉 𝑦 is equal to 𝑉 multiplied by the sine of the launch angle above the horizontal, 𝜃. Substituting this back into our expression for the time of flight of the projectile, we get 𝑇 equals two 𝑉 sin 𝜃 divided by 𝑔.

All that’s left to do is for us to substitute our values of 𝑉, 𝜃, and 𝑔 into this equation to calculate the time of flight of the projectile. The question tells us that the projectile’s initial speed is 25 meters per second, so 𝑉 is equal to 25 meters per second. The question also tells us that the launch angle of the projectile is 48 degrees above the horizontal, so 𝜃 is equal to 48 degrees. Finally, 𝑔 is equal to 9.81 meters per second squared. 𝑔 and 𝑉 are in SI units, so we don’t need to worry about converting either of them.

So, 𝑇 equals two multiplied by 25 meters per second multiplied by the sin of 48 degrees divided by 9.81 meters per second squared. To one decimal place, 𝑇 equals 3.8 seconds. The time between the projectile leaving the ground and returning to the ground at the same height that it was launched from is 3.8 seconds.