# Vertical Projectile Motion Problem

• TheRedDevil18
In summary, the two balls thrown vertically downwards and upwards at the same speed hit the ground at different times and have a difference in flight time of 3 seconds.
TheRedDevil18

## Homework Statement

Two students are on a balcony 19.6m above the ground. One guy throws a ball vertically downwards at 14.7m/s; at the same time the other guy throws a second ball vertically upwards at the same speed. The second ball just misses the balcony on the way down.

2.3.1) What is the difference in the balls flight time?
2.3.2) What is the velocity of each ball as they strike the ground?
2.3.3) How far apart are the balls 0.8s after they are thrown?

## Homework Equations

vf^2 = vi^2 + 2a*y
delta y = vi*t + 1/2*a*t^2
delta y = vf+vi/2 * t
vf = vi + a*t

## The Attempt at a Solution

2.3.1) For the ball thrown upwards I decided to first find the maximum height it reached. I used the first equation from the list and got an answer of 11.03m. I then used the third equation to find the time and got 1.5s (another 1.5s for coming down=3s). I then worked out the time for the ball thrown downwards to be 1s. My final answer for the difference in flight time is 4-1=3s.

2.3.2) Each ball would strike the ground at the same speed.
vf = vi+a*t
= 14.7+9.8*1
= 144.06m/s

2.3.3) Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.

TheRedDevil18 said:
1. 2.3.3) Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.

Check the number in red above! Otherwise all is good.

I think the first ball should be 14.896 and the second ball 8.624. Is that my mistake?

TheRedDevil18 said:
1.

= 14.7+9.8*1 = 144.06m/s

Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.

Check the items in red above!

Ok I am a bit confused, but this is how I got 14.896m:
delta y = 14.7*0.8 + 1/2(9.8)*0.8^2
= 14.896m

Is there something I did wrong?

TheRedDevil18 said:
Ok I am a bit confused, but this is how I got 14.896m:
delta y = 14.7*0.8 + 1/2(9.8)*0.8^2
= 14.896m

Is there something I did wrong?

Should that number be + or -?

If the answer should be negative, Why?

TheRedDevil18 said:
If the answer should be negative, Why?

Because when you came up with y1 = +8.624m you defined y = 0 to be the balcony height, so anything below the balcony is negative. It should also be obvious that the second ball will be lower in height than the first, not higher as you had it.

So y1 = vi*t - gt^2/2 = 8.624m (ball tossed up but g is down)
y2 = -vi*t - gt^2/2 = -14.896m (ball tossed down & g is down)
y1 - y2 = 2vi*t = you fill this one in!

Just keep track of signs & you'll be fine.

So the final answer is 23.52m apart?

TheRedDevil18 said:
So the final answer is 23.52m apart?

Yep.

## 1. What is vertical projectile motion?

Vertical projectile motion is a type of motion in which an object is launched into the air and moves only in the vertical direction due to the force of gravity. This type of motion is often seen in activities such as throwing a ball or jumping off a diving board.

## 2. What factors affect vertical projectile motion?

The factors that affect vertical projectile motion include the initial velocity, the angle of launch, the mass of the object, and the force of gravity. These factors can impact the height and distance traveled by the object, as well as the time it takes to reach its highest point and return to the ground.

## 3. How is vertical projectile motion calculated?

To calculate vertical projectile motion, the equations of motion can be used, which take into account the initial velocity, angle of launch, and gravitational force. These equations can be solved for variables such as time, velocity, and displacement, and can provide information about the object's motion at any point in time.

## 4. What is the maximum height of a vertically launched projectile?

The maximum height of a vertically launched projectile is determined by the initial velocity and the angle of launch. The higher the initial velocity and the closer the angle of launch is to 90 degrees (directly upwards), the higher the maximum height will be. This can be calculated using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the displacement (in this case, the maximum height).

## 5. How does air resistance affect vertical projectile motion?

Air resistance can affect vertical projectile motion by slowing down the object's velocity and changing its trajectory. This is because the force of air resistance acts in the opposite direction of motion, causing the object to lose speed and eventually fall to the ground. The impact of air resistance can vary depending on the size, shape, and density of the object, as well as the air density and other environmental factors.

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