Calculating the Velocity of Snowballs Thrown at 40°

  • Thread starter keweezz
  • Start date
  • Tags
    Velocity
In summary: I just thought it might be easier to visualize with components.In summary, two individuals throw snowballs from a roof, one downward at 40° below the horizontal and the other upward at 40° above the horizontal. Both snowballs have an initial velocity of 13 m/s and are thrown from an elevation of 14 m. When they reach a height of 5.0 m, the vertical component of the velocity is determined to be 15.7 m/s for both snowballs. The horizontal component remains unchanged. Using the formula Vf^2 = Vi^2 + 2ax, the vertical component is found by using the distance of +9 (in the direction of velocity and acceleration) and the acceleration due to
  • #1
keweezz
12
0
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 14 m with the same initial speed of 13 m/s, but in different directions. You through your snowball downward, at 40° below the horizontal; your friend throws her snowball upward, at 40° above the horizontal. What is the speed of each ball when it is 5.0 m above the ground? (Neglect air resistance.)


m/s (your snowball)
m/s (your friend's snowball)

15.7 m/s is what I keep getting... I did 13m/s sin 30 = 8.3 and.. i used that for Vf^2 = vi^2 + 2ax and used -9 for x and -9.81 for a...

well 15.7 is not right... any thought?
 
Physics news on Phys.org
  • #2
Why did you use -9 for x? First, find how long each snowball will be in the air.
 
  • #3
sin 30 or 40?
 
  • #4
keweezz said:
You and a friend stand on a snow-covered roof. You both throw snowballs from an elevation of 14 m with the same initial speed of 13 m/s, but in different directions. You through your snowball downward, at 40° below the horizontal; your friend throws her snowball upward, at 40° above the horizontal. What is the speed of each ball when it is 5.0 m above the ground? (Neglect air resistance.)


m/s (your snowball)
m/s (your friend's snowball)

15.7 m/s is what I keep getting... I did 13m/s sin 30 = 8.3 and.. i used that for Vf^2 = vi^2 + 2ax and used -9 for x and -9.81 for a...

well 15.7 is not right... any thought?

Remember velocity is composed of 2 components. What is the horizontal (x) component of velocity. That should stay the same with no air resistance. Then you have the vertical component to deal with. That is where you need to take into account the effect of gravity.
 
  • #5
i used -9 because that was the difference between 5-14? and its sin 40 :O

so i find t and use Vi+at=Vf?

i find t using x=Vit+1/2a(t)^2?
 
  • #6
keweezz said:
i used -9 because that was the difference between 5-14? and its sin 40 :O

so i find t and use Vi+at=Vf?

i find t using x=Vit+1/2a(t)^2?

Yes that will give you the vertical velocity component. But the distance you should use is +9 as that is in the direction your velocity and acceleration is.

Then you need to combine that with the horizontal component. (Velocity is a vector. You add the two vectors (x,y) together.)

Edit: Btw there is nothing wrong with using the Vf2 = Vo2 +2aΔx you started out with.
 
Last edited:

FAQ: Calculating the Velocity of Snowballs Thrown at 40°

What is the formula for calculating the velocity of a snowball thrown at 40°?

The formula for calculating the velocity of a snowball thrown at 40° is v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the throw.

What units should be used when calculating the velocity of a snowball thrown at 40°?

The units used when calculating the velocity of a snowball thrown at 40° should be meters per second (m/s) for velocity, meters (m) for height, and meters per second squared (m/s²) for acceleration due to gravity.

How does the angle of the throw affect the velocity of a snowball?

The angle of the throw does not affect the velocity of a snowball. The velocity is determined by the initial height and acceleration due to gravity, not the angle of the throw.

What is the average velocity of a snowball thrown at 40°?

The average velocity of a snowball thrown at 40° will vary depending on the initial height and acceleration due to gravity. It can be calculated using the formula v = √(2gh), where g is the acceleration due to gravity and h is the average height of the throw.

Can the velocity of a snowball thrown at 40° be negative?

Yes, the velocity of a snowball thrown at 40° can be negative if the initial height is below the starting point. This indicates that the snowball is moving downwards instead of upwards.

Similar threads

Back
Top