Question Video: Calculating the Velocity of a Projectile at Multiple Points along Its Trajectory Physics • 9th Grade

A ball is launched vertically upward and then returns to the ground. The change of the vertical displacement of the ball with time is shown on the graph. The maximum vertically upward displacement of the ball from its launch point is 5 m. What is the initial vertical velocity of the ball? Give your answer to the nearest meter per second? What is the speed of the ball at point P, where its upward vertical displacement from its launch point is 4.5 m. Give your answer to one decimal place.

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Video Transcript

A ball is launched vertically upward and then returns to the ground. The change of the vertical displacement of the ball with time is shown on the graph. The maximum vertically upward displacement of the ball from its launch point is five meters. What is the initial vertical velocity of the ball? Give your answer to the nearest meter per second? What is the speed of the ball at point P, where its upward vertical displacement from its launch point is 4.5 meters. Give your answer to one decimal place.

Letโ€™s start with the first part of this question. In this question, we have a ball which we are told is launched vertically upward. So it has some initial vertical velocity that we will call ๐‘‰ sub i. During the ballโ€™s motion, the only force acting on it is gravity, which acts vertically downwards and has a magnitude that is equal to the mass of the ball, which we will call ๐‘š, multiplied by the acceleration due to gravity, which is ๐‘”. This downward force on the ball causes its vertical velocity to decrease over time, meaning it will rise through the air until it reaches a maximum vertically upward displacement. And at this point the ball will have no vertical velocity. So we can write ๐‘‰ is equal to zero.

After this point, the vertical velocity of the ball will become negative, causing it to travel downwards through the air until it reaches the ground. The vertical displacement of the ball from its launch point is shown on the graph, where the vertical axis of the graph shows its displacement and the horizontal axis of the graph shows time. The question actually tells us that the maximum vertically upward displacement of the ball from its launch point is five meters. So before we go any further, letโ€™s add this to the graph.

This first part of the question asked us to work out the initial vertical velocity of the ball. So we must calculate ๐‘‰ sub i. To answer this part of the question, we will use an equation that relates to the velocity and displacement of an object undergoing constant acceleration. The equation states that the final velocity of the object squared is equal to its initial velocity squared plus two multiplied by the acceleration the object is experiencing multiplied by its displacement.

If we take the launch point of the ball as its initial point and the point at which the ball reaches its maximum altitude as its final point, then we know that the final velocity of the ball is zero meters per second. The acceleration that the ball is experiencing is negative ๐‘”. And this is negative because gravity acts downwards which is the opposite direction of positive vertical displacement. We also know that the value of ๐‘” is 9.8 meters per second squared. So the acceleration that the ball is experiencing is negative 9.8 meters per second squared. Finally, weโ€™re told that the vertical displacement of the ball at this final point is five meters.

Now, letโ€™s clear some space on the right and see how we can use all of this to answer the question. Weโ€™ll start by rearranging this equation for the initial velocity of the ball. Subtracting minus two ๐‘Ž๐‘  from both sides, we see that the two ๐‘Ž๐‘ โ€™s on the right cancel. Then taking the square root of both sides, we see that the square root of ๐‘‰ sub i squared is just ๐‘‰ sub i. So this gives us our equation for the initial velocity of the ball. Writing this bit more neatly, ๐‘‰ sub i is equal to the square root of ๐‘‰ sub f squared minus two ๐‘Ž๐‘ . Now all we have to do is substitute our known value of ๐‘‰ sub f, ๐‘Ž, and ๐‘  into this equation. And these are all in SI units, so we donโ€™t need to worry about converting them before we continue.

Substituting them in, we get ๐‘‰ sub i is equal to the square root of zero meters per second squared minus two multiplied by minus 9.8 meters per second squared multiplied by five meters. We can immediately see that this zero meters per second makes no contribution towards this at all. Then, if we take the negative outside of the brackets, it cancels with the other negative. Expanding the brackets, we get ๐‘‰ i is equal to the square root of 98 meters squared per second squared. Now these units may look a little bit odd, but when we take the square root of them, we see that they just give us meters per second, which are the units weโ€™d expect for velocity.

Evaluating this expression gives ๐‘‰ i is equal to 9.90 meters per second. However, this part of the question asks us to give our answer to the nearest meter per second. 9.90 meters per second to the nearest meter per second is just 10 meters per second. So ๐‘‰ sub i is equal to 10 meters per second. Therefore, the initial vertical velocity of the ball is 10 meters per second to the nearest meter per second.

Now, letโ€™s take a look at the second part of this question.

The second part of the question asks us, what is the speed of the ball at point P, where its upward vertical displacement from its launch point is 4.5 meters? Give your answer to one decimal place.

So we now want to work out the speed of the ball at a different points along its trajectory. And we could have two options of how to work this out. Our first option is to set point P as our new final point in our calculations. And we can use the same equation as before with our calculated value of the initial vertical velocity of the ball and the new displacement that we are told is 4.5 meters to work out the speed of the ball at point P.

Our second option is actually to set point P as the initial point in our calculations. And we note that the relative vertical displacement between the initial and final points would now be 0.5 meters. Then, we can use the exact same calculations as we did for the first part of the question to answer the second part. Weโ€™ll use this second option. And letโ€™s remind ourselves that the final velocity of the ball is zero meters per second, the acceleration it experiences is negative 9.8 meters per second squared, and the displacement of the ball is now the relative displacement between its initial and final points, which we know is equal to 0.5 meters.

Given all of this information, we need to find the speed of the ball at point P. Since this is the initial point in our calculation, we need to find the magnitude of the velocity of the ball at this initial point. So we must calculate ๐‘‰ i and then take the magnitude of this, and we will have our answer.

Just like in the first part of this question, we will start with this equation here. And once again, we will rearrange for ๐‘‰ sub i. We will start by subtracting two ๐‘Ž๐‘  from both sides, where we see that these two terms on the right cancel. Then, we will take the square root of both sides. And we see that on the right, the square root of ๐‘‰ sub i squared is just ๐‘‰ sub i. And this gives us our equation for the initial velocity of the ball. Writing this a bit more neatly, ๐‘‰ sub i is equal to the square root of ๐‘‰ sub f squared minus two ๐‘Ž๐‘ .

Now all we have to do is substitute our values for ๐‘‰ sub f, ๐‘Ž, and ๐‘  into this equation. And we should once again note that these are all in SI units, so we donโ€™t need to convert any of them before we continue. Substituting these in gives ๐‘‰ sub i is equal to the square root of zero meters per second squared minus two multiplied by negative 9.8 meters per second squared multiplied by 0.5 meters. Evaluating this expression gives us a value of ๐‘‰ i is equal to 3.13 meters per second. So the velocity of the ball at point P is equal to 3.13 meters per second. And because this is positive, this has the same value as the ballโ€™s speed. So we can write that the speed is equal to 3.13 meters per second.

Finally, the question asked for the answer to one decimal place. 3.13 to one decimal place is just 3.1. So the speed of the ball at point P, where its upward vertical displacement from its launch point is 4.5 meters, is 3.1 meters per second to one decimal place.

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