Video Transcript
A projectile is fired at an angle of 55 degrees above the horizontal and has a
maximum upward vertical displacement from its launch position of 7.2 meters. What is the initial speed of the projectile? Give your answer to the nearest meter per second.
In this question, we have a projectile that is fired with some initial speed that we
will call 𝑉. It is launched at an angle above the horizontal that we will call 𝜃. And the question tells us that this is an angle of 55 degrees above the
horizontal. So we can make a note that 𝜃 is equal to 55 degrees.
Now, because this is a projectile, the only force that is acting on it is gravity,
and this acts vertically downwards and has a magnitude of the mass of the projectile
that we will call 𝑚 multiplied by the acceleration due to gravity, which is 𝑔. This downward force causes the projectile’s trajectory to be curved. And if we look at the projectile’s vertical velocity, we see that it decreases
throughout its motion, meaning that at some point, the projectile will have no
vertical velocity. After this, it will start moving downwards.
This point where the vertical velocity of the projectile is zero is the point at
which the projectile has its maximum upward vertical displacement, also known as its
maximum altitude. We will call this maximum upward vertical displacement from its launch position
ℎ. And we are told in the question that this has a value of 7.2 meters. So we can make a note that ℎ is equal to 7.2 meters. The question wants us to work out the initial speed of the projectile given this
information, so we must calculate 𝑉. Before we go any further, there’s another value that we should take a note of, and
that’s the acceleration due to gravity. This is 𝑔, and it has a value of 9.8 meters per second squared.
To answer this question, we will use an equation that describes the motion of an
object undergoing constant acceleration. The equation states that the final velocity of the object squared is equal to the
initial velocity of the object squared plus two multiplied by the acceleration the
object is experiencing multiplied by the object’s displacement. In this question, we will apply this to the vertical motion of the projectile, with
the beginning of the motion being our initial point and the position where the
projectile is at its maximum altitude as our final point. We already know that the projectile has no vertical velocity when it is at this point
of maximum altitude, so 𝑉 f is equal to zero.
The initial vertical velocity of the projectile can be calculated by looking at a
diagram of its initial speed, which we can see has a horizontal component and a
vertical component. And we will call the vertical component 𝑉 y. We can see that these form a right angle triangle, so we can see that 𝑉 𝑦is equal
to 𝑉 multiplied by the sin of 𝜃. The initial vertical velocity of the projectile is equal to its initial speed
multiplied by the sin of its launch angle above the horizontal. And we can write this into our equation. The acceleration that the projectile is experiencing is equal to 𝑔, but it acts
downwards. So we add a negative sign.
Finally, the vertical displacement of the projectile is equal to ℎ. Writing this a bit more neatly, zero is equal to 𝑉 sin 𝜃 all squared minus two
𝑔ℎ. And we know the value of 𝜃, 𝑔, and ℎ, so we just need to rearrange this equation
for 𝑉. First, we will add two 𝑔ℎ to both sides, where we can see that these terms on the
right cancel. And two 𝑔ℎ plus zero is just equal to two 𝑔ℎ. Next, we will take the square root of both sides, which actually just leaves us with
𝑉 sin 𝜃 on the right. Next, we will divide both sides by sin 𝜃, where we see that the sin 𝜃s on the right
cancel. And this gives us our expression for 𝑉, the initial speed of the projectile. Writing this a bit more clearly, 𝑉 is equal to the square root of two 𝑔ℎ divided by
the sin of 𝜃.
Our final step is to substitute our known values of 𝜃, ℎ, and 𝑔 into this
equation. Before we continue, we should check our units. ℎ and 𝑔 are both in SI units, and the sin of 𝜃 doesn’t have any units, so we
actually don’t need to convert any of these before continuing. Substituting these in, we get 𝑉 is equal to the square root of two multiplied by 9.8
meters per second squared multiplied by 7.2 meters divided by the sin of 55
degrees. Evaluating this gives us 𝑉 is equal to 14.5 meters per second. But the question actually asks us to give our answer to the nearest meter per
second. So 𝑉 is equal to 15 meters per second, and this is our projectile’s initial speed
and is the answer to our question. The initial speed of the projectile is 15 meters per second to the nearest meter per
second.
