Projectile Motion: Solving for Time & Velocity with No Air Resistance

  • Thread starter Crusaderking1
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In summary: Is this right?Hey!I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2Can you please explain how you arrived at substracting your desired height from zero?It should be, just 10.7, nothing else.DanielRemember: Vy = Voy - gtI think you should revise your Vy => Although for this problem it is not necessary
  • #1
Crusaderking1
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Homework Statement



A major leaguer hits a baseball so that it leaves the bat at a speed of 30.1 m/s and at an angle of 37.2 degrees above the horizontal. You can ignore air resistance.

a) At what two times is the baseball at a height of 10.7 m above the point at which it left the bat?

Voy = 18.2 m/s (30.1 sin37.2)
Vox= 24.0 m/s (30.1 cos37.2)

ax=0
ay= -9.8 m/s^2

yo = 0
xo = 0

vy= -18.2 m/s
xy= 24.0 m/s

Homework Equations



any constant acceleration equations.

The Attempt at a Solution



I received Ttotal = 3.71 seconds which I think is right.

I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2

which is -4.9t^2 + 18.2t + 10.7

I did the quadratic equation but received the times -0.516 and 4.23, and added together equal my total time of 3.71, however, I don't understand how there is a negative number.

Are these the right times?

P.S (this is a 7 part question, so I might need help after this question).

Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

I'm guessing this means use .732 seconds for time.
I used the equation: vy=voy +ayt

Vy= 18.2+(-9.8)(.732)

=11.03 m/s.

Is this right?
 
Last edited:
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  • #2
Hey!
I set up the equation to find time by doing 0-10.7 = 18.2t + .5(-9.8)t^2
Can you please explain how you arrived at substracting your desired height from zero?
It should be, just 10.7, nothing else.
Daniel
 
  • #3
Remember: Vy = Voy - gt

I think you should revise your Vy => Although for this problem it is not necessary

use this
Yf = Yo + Voyt - .5gt^2

just think where you want your Yf to be and solve for t
 
  • #4
danielakkerma said:
Hey!

Can you please explain how you arrived at substracting your desired height from zero?
It should be, just 10.7, nothing else.
Daniel

I received the new times of 0.732 and 2.98 which sounds much better!

I should have had 10.7-0, not 0-10.7.
 
  • #5
cd.riter said:
Remember: Vy = Voy - gt

I think you should revise your Vy => Although for this problem it is not necessary

use this
Yf = Yo + Voyt - .5gt^2

just think where you want your Yf to be and solve for t

sounds much easier than my method! thanks.
 
  • #6
Always look for that blasted minus sign!
Well done!
Daniel
 
  • #7
danielakkerma said:
Always look for that blasted minus sign!
Well done!
Daniel

thanks!
 
  • #8
Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a).

I'm guessing this means use .732 seconds for time.
I used the equation: vy=voy +ayt

Vy= 18.2+(-9.8)(.732)

=11.03 m/s.

Nevermind, I know this is right.
 
Last edited:

1. What is a projectile?

A projectile is an object that is launched into the air and moves along a curved path due to the force of gravity acting on it.

2. What is the formula for calculating the trajectory of a projectile?

The formula for calculating the trajectory of a projectile is y = y0 + v0t + 1/2at2, where y is the vertical position, y0 is the initial height, v0 is the initial velocity, t is time, and a is the acceleration due to gravity.

3. How does air resistance affect the trajectory of a projectile?

Air resistance, or drag, can affect the trajectory of a projectile by slowing it down and altering its path. This is because air resistance acts in the opposite direction of the projectile's motion, reducing its speed and causing it to deviate from its expected trajectory.

4. How does the angle of launch affect the range of a projectile?

The angle of launch, or the angle at which the projectile is launched into the air, can greatly affect its range. The optimal angle for maximum range is 45 degrees, as this allows for the perfect balance between horizontal and vertical components of the projectile's velocity.

5. What is the maximum height reached by a projectile?

The maximum height reached by a projectile is determined by its initial velocity and the acceleration due to gravity. This height can be calculated using the formula h = (v02sin2θ)/2g, where v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

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