Question Video: Calculating the Launch Angle of a Projectile | Nagwa Question Video: Calculating the Launch Angle of a Projectile | Nagwa

Question Video: Calculating the Launch Angle of a Projectile Physics • First Year of Secondary School

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A ball was thrown from the ground at an angle πœƒ with a velocity 𝑣. The ball reached a maximum height of 8 m and a maximum range of 15 m. Calculate the value of πœƒ. Let 𝑔 = 10 m/sΒ².

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Video Transcript

A ball was thrown from the ground at an angle πœƒ with a velocity 𝑣. The ball reached a maximum height of eight meters and a maximum range of 15 meters. Calculate the value of πœƒ. Let 𝑔 equal 10 meters per second squared. (A) 82.4 degrees, (B) 64.9 degrees, (C) 46.8 degrees, and (D) 28.1 degrees.

In this question, we’ve been told that a ball is thrown from the ground at an angle πœƒ to the horizontal. We also know that the maximum height of the ball’s trajectory is eight meters, and the horizontal distance the ball travels between when it’s thrown and when it reaches the ground again is 15 meters. Based on this information, we need to calculate the angle πœƒ between the horizontal and the direction of the ball’s initial velocity 𝑣.

To calculate πœƒ, let’s begin by saying that the ball takes a time capital 𝑇 to complete its trajectory. Since we will be ignoring any air resistance acting on the ball, we can assume the trajectory is symmetric about the midpoint and the ball is found at its maximum height, exactly halfway along its journey, at a time of capital 𝑇 over two.

Next, it’s worth noting that the acceleration due to gravity is 10 meters per second squared. This will only act in the vertical direction, since the gravitational force on the ball acts down towards the ground. In other words, if the horizontal direction is labeled π‘₯ and the vertical direction is labeled 𝑦, we can say that the acceleration in the 𝑦-direction, π‘Ž sub 𝑦, is equal to negative 10 meters per second squared, or negative 𝑔 as it’s been given to us in the question.

The reason this quantity is negative is because we implicitly assumed that any motion in the upward direction is positive when we labeled the maximum height of the ball as being positive eight meters. The ball moved upward to get to a maximum vertical displacement of eight meters, so a downward acceleration will have a negative value. We’ve also been told in the question that the ball is thrown with a velocity 𝑣. We can find the horizontal and vertical components of this velocity, so we can deal with the horizontal and vertical motion of the ball separately.

Let’s recall that the ball’s initial horizontal velocity, which we will call 𝑣 sub π‘₯, is equal to 𝑣 multiplied by the cos of the angle πœƒ. Similarly, the initial vertical velocity component, 𝑣 sub 𝑦, is equal to 𝑣 multiplied by the sin of πœƒ. The reason we do this is because we can now study the horizontal and vertical motion of the ball individually, which will make our calculations simpler to deal with.

For example, let’s recall that for an object moving at a constant velocity, the velocity of that object is equal to the displacement of the object divided by the time taken to travel this displacement. We can apply this equation to the horizontal motion of the ball because in this direction, the velocity of the ball is indeed constant.

Hence, clearing some space to work, we find that the horizontal velocity of our ball, 𝑣 times the cos of πœƒ, is equal to the horizontal displacement 𝑠 sub π‘₯ divided by the time taken to move this distance, which we’ve called capital 𝑇. Now we know the value of 𝑠 sub π‘₯ β€” it’s 15 meters β€” but we won’t substitute this in just yet.

Next, let’s apply some kinematic equations to the horizontal and vertical motion of the ball. Let’s recall that the kinematic equations can only be applied to objects moving with a constant acceleration. Our ball fits this criterion, as it has a constant downward acceleration due to gravity. The first kinematic equation we can recall is this one. The final velocity of an object, 𝑣 sub f, is equal to the initial velocity, 𝑣 sub i, plus the acceleration, π‘Ž, multiplied by the time for which it moves 𝑑.

We can apply this to the vertical motion of the ball between when it is thrown and when it reaches its maximum height. The final vertical velocity of the ball in this scenario is actually zero. When the ball is at its maximum height, it must have exactly zero vertical velocity. Because if this was not true, then the ball would either continue moving upward or not have got this high, and the maximum height would be something else. So, 𝑣 sub f 𝑦 is equal to the initial vertical velocity, which is 𝑣 times the sin of πœƒ, plus the acceleration, which is negative 𝑔, multiplied by the time for which the ball moves.

Since we’re only studying the motion up until the ball reaches the maximum height, this time is going to be capital 𝑇 divided by two, half the time for the entire motion. We can now rearrange these two equations we’ve found so we can solve for 𝑣 in both cases. For this equation, where recall 𝑣 sub f 𝑦 is zero, we find 𝑣 is equal to 𝑔 times capital 𝑇 divided by two times the sin of πœƒ. And the first equation can be rearranged so it reads 𝑣 is equal to 𝑠 sub π‘₯ divided by capital 𝑇 times the cos of πœƒ.

Now, we can equate the two right-hand sides of these equations as they are both equal to 𝑣. When we do this, we can rearrange once again. So, our equation reads 𝑠 sub π‘₯ multiplied by the sin of πœƒ divided by the cos of πœƒ is equal to 𝑔 times capital 𝑇 squared all divided by two. The reason we rearrange like this is because we can recall that sin πœƒ divided by cos πœƒ is equal to tan πœƒ.

Next, we can apply another kinematic equation to the vertical motion of the ball. This equation tells us that the displacement of an object is equal to the initial velocity multiplied by the time of travel plus one-half times the acceleration times the time squared.

For the ball falling from its maximum height back to ground, the vertical displacement is negative 𝑠 sub 𝑦, the initial vertical velocity is zero, the acceleration is negative 𝑔, and the time is capital 𝑇 divided by two. Canceling out the term which is zero, due to the initial velocity being zero, and canceling the negative signs on both remaining terms, we find that 𝑠 sub 𝑦 is equal to one-half times 𝑔 times capital 𝑇 squared over four.

Another way to write the right-hand side of this equation is as one-fourth times the quantity one-half multiplied by 𝑔 times capital 𝑇 squared. The reason we do this is because our equation from earlier also has the quantity one-half 𝑔 times capital 𝑇 squared as its right-hand side. This means we can equate the two left-hand sides if we remember to account for this factor of one-fourth. We find that 𝑠 sub 𝑦 is equal to one-fourth multiplied by 𝑠 sub π‘₯ times the tan of πœƒ.

At this point, we know all the quantities in this equation apart from πœƒ, so we can rearrange to solve for it. We find that tan of πœƒ is equal to four times 𝑠 sub 𝑦 divided by 𝑠 sub π‘₯. So, πœƒ is equal to the inverse tan of this quantity. We finally substitute in 𝑠 sub 𝑦 is equal to eight meters and 𝑠 sub π‘₯ is equal to 15 meters. The units of meters in the numerator and denominator cancel, which is great because we need a unitless number in order to be able to find the inverse tangent of it. When we evaluate this expression, we find πœƒ to be equal to 64.88 and so on degrees. To one decimal place, this becomes 64.9 degrees.

Looking back at our answer options, we see this corresponds to option (B). Therefore, we’ve found that the angle between the ground and the initial velocity of the ball is 64.9 degrees.

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