Projectile motion ball thrown down vertically

In summary, the ball thrown vertically downward from the top of a building with a velocity of 30 ft will have a speed of -94 ft/s after falling 2 seconds and will fall 124 ft in 2 seconds. Its velocity after 30 ft will be -53.04 ft/s. If the ball was released at 120 ft above the ground, it will strike the ground in 2.74 seconds.
  • #1
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Homework Statement


A ball is thrown vertically downward with a velocity of 30 ft from the top of a building.

Find

a.)What it's speed will be after falling 2 seconds
b.)How far it will fall in 2 seconds
c.)What will be its velocity after 30 ft
d.)If the ball was released at 120 ft above the ground, in how many seconds will it strike the ground.


Homework Equations



For a I used the equation Vy = Vyo - gt
For b I used the equation ΔY=Vyo(t) - 1/2gt^2
For c I used the equation ΔY=Vyo(t) - 1/2gt^2 in order to find t, or the time it takes to move 30 ft and I got .72 s. Then I put that into the first equation of Vy=Vyo - gt.
For d I used ΔY=Vyo(t) - 1/2gt^2.

The Attempt at a Solution



For a I got -94 ft/s.
For b I got -124 ft/s.
For c I got 53.04 ft/s.
For d I got 2.74 seconds

I have a few questions. First, can someone check if these answers are correct? My professor did not provide an answer sheet for the exam review. Also I was only able to get a proper answer for d if I made the 120 ft negative. If the ball was released at 120 ft above the ground wouldn't it be positive? Also, c gave me some trouble, if it is incorrect, could someone please provide an in-depth explanation? Thank you so much.
 
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  • #2
I think your answers for (a) and (b) are correct.

For (c), it should be -53.04ft/s.
Instead of working out the time first, you can just use 1 equation :
V2 = V02 - 2gΔy

For (d), you release the ball at the origin of your coordinate system, so the ground is at y=-120ft.
From (b) you know the ball falls 124ft in 2s, so to fall 120ft will be just under 2s, so your answer is not correct here.
 
  • #3
You should be careful with the units - "30ft" is not a velocity, and "-124ft/s" is not a length.
 
  • #4
ap123 said:
I think your answers for (a) and (b) are correct.

For (c), it should be -53.04ft/s.
Instead of working out the time first, you can just use 1 equation :
V2 = V02 - 2gΔy

For (d), you release the ball at the origin of your coordinate system, so the ground is at y=-120ft.
From (b) you know the ball falls 124ft in 2s, so to fall 120ft will be just under 2s, so your answer is not correct here.

Well the question for d did not specify an initial velocity, it just said released, so I think it's right.

mfb said:
You should be careful with the units - "30ft" is not a velocity, and "-124ft/s" is not a length.

Oh right, I better not make that mistake on the exam haha.
 
  • #5
For part (d), yes you're correct.
I didn't notice it started from rest.
 
  • #6
ap123 said:
For part (d), yes you're correct.
I didn't notice it started from rest.

Cool, thank you very much for your assistance.
 

FAQ: Projectile motion ball thrown down vertically

1. What is projectile motion?

Projectile motion is the motion of an object that is thrown, launched, or dropped and moves along a curved path due to the influence of gravity.

2. How does gravity affect projectile motion?

Gravity is the force that pulls objects towards the center of the Earth. In projectile motion, gravity acts on the object and causes it to accelerate downward, changing the direction of its motion.

3. What is the initial velocity in projectile motion?

The initial velocity in projectile motion is the speed and direction at which the object is launched. It determines the path and range of the projectile.

4. What is the trajectory of a projectile?

The trajectory of a projectile is the path that it follows through the air. It is a curved path that is influenced by the initial velocity, gravity, and air resistance.

5. How can the maximum height and range of a projectile be calculated?

The maximum height and range of a projectile can be calculated using the equations of motion for projectile motion. The maximum height can be found using the formula h = (v2sin2θ)/2g and the range can be found using the formula R = (v2sin2θ)/g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

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