Projectile motion from edge of building

[J89]

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..

 

[LowlyPion]

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.

 

[nlightner]

I would like to clarify that the given equation is not entirely correct for this scenario. The correct equation for projectile motion is:

y = y0 + v0y*t - 1/2*g*t^2

Where:
y = vertical position at time t
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (9.8 m/s^2)
t = time

To solve for the y component of the projectile's position at t = 2.4 seconds, we need to first find the initial vertical velocity (v0y) of the projectile. We can do this by breaking down the initial velocity (v0) into its horizontal and vertical components using trigonometry.

v0x = v0*cos(45) = 10*cos(45) = 7.07 m/s
v0y = v0*sin(45) = 10*sin(45) = 7.07 m/s

Now, we can substitute these values into the equation:

y = y0 + v0y*t - 1/2*g*t^2
y = 0 + 7.07*2.4 - 1/2*9.8*(2.4)^2
y = 12.6 meters

Therefore, the y component of the projectile's position at t = 2.4 seconds is 12.6 meters.

To find the maximum height the projectile reaches, we can use the fact that the vertical velocity at the highest point is 0 m/s. We can find this point by setting v0y*t - 1/2*g*t^2 = 0 and solving for t.

0 = 7.07*t - 1/2*9.8*t^2
t = 1.43 seconds

Now, we can substitute this value into the equation for y to find the maximum height:

y = 0 + 7.07*1.43 - 1/2*9.8*(1.43)^2
y = 5.06 meters

Therefore, the maximum height the projectile reaches is 5.06 meters.

I would also like to mention that the given drawing is not accurately scaled, as the vertical distance traveled by the projectile is much greater than the horizontal distance. In reality, the projectile would reach a much higher height before landing 35.4