Projectile motion from edge of building

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a projectile fired from the edge of a building with an initial velocity of 10 m/s at a 45-degree angle. The projectile lands 35.4 meters from the base of the building. Key equations used include the vertical motion equation Y = h + Vy*t - 1/2*g*t², where Vy is the vertical component of the initial velocity calculated as 7.07 m/s. The height of the building (h) can be determined using the time of flight derived from the horizontal distance traveled.

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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..

Homework Equations


Vsina0 + -1/2gt^2

The Attempt at a Solution


Used the given equation for both parts but got them wrong..
 

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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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