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Projectile motion

  • Thread starter J89
  • Start date

J89

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1. Homework Statement
A projectile is fired from the edige of a building with initial velocity of v0=10 m/s at an angle of 45 degrees. The rojectile rises and then lands oln ground at point P as shown, which is 35.4 meters away from the base of building. A) Find the Y component of the projectile's position at t =2.4 seconds and the maximum height the projectile reaches. Drawing is attached too..



2. Homework Equations
Vsina0 + -1/2gt^2



3. The Attempt at a Solution
Used the given equation for both parts but got them wrong..
 

Attachments

LowlyPion

Homework Helper
3,079
4
You know vertical velocity initially.

10 * .707 = 7.07 m/s

But your equation is wrong.

Y = h + Vy*t - 1/2*g*t2

You need to figure h the height of the building.

You can determine that from the x-distance at impact

35.4/Vx = 35.5 / 7.07 = total time

Since at total time that tells you where the object reaches 0 in the equation, that should tell you h. With h, you can figure Y at 2.4 sec.

Figure the time to max height and that gives you the Y of max height, all from the same equation for Y.
 

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