An object is launched at an angle of 24 degrees above the horizontal at a velocity of 33.7 m/s from the top of a building which is 110.8 meters above the ground. When the object lands on the ground, what is the horizontal displacement from the base of the building in meters?


Hint: First find the final velocity of the object in the y direction when it hits the ground using the given information above. Then use this final velocity in the y direction to find the time it takes to achieve this velocity in the y direction using a simple projectile motion equation. With this time it should be straight forward to find the horizontal displacement.

Vo = 33.7m/s[24o]

Xo = 33.7*Cos24 = 28 m/s.
Yo = 33.7*Sin24 = 13.7 m/s.

Tr = -Yo/g = -13.7/-9.8 = 1.40 s. = Rise time.

h max = -Yo^2/2g = -(13.7^2)/-19.6 = 9.58 m. Above the bldg.

0.5g*t^2 = 9.58 + 110.8 = 120.4 m
4.9*t^2 = 120.4
t^2 = 24.57
Tf = 4.96 s. = Fall time.

Dx = Xo*(Tr+Tf) = 28 * (1.40+4.96) = 176.1 m.

To solve for the horizontal displacement, we first need to find the final velocity of the object in the y direction when it hits the ground. We can do this by using the given information and applying the equations of projectile motion.

Given:
Initial velocity (v0) = 33.7 m/s
Launch angle (θ) = 24 degrees
Height of the building (h) = 110.8 meters

Step 1: Find the final velocity in the y direction (vfy).
The final velocity in the y direction will be zero when the object hits the ground. We can use the vertical motion equation:

vfy^2 = v0y^2 + 2ay * Δy

Where:
vfy = final velocity in the y direction (which is zero in this case)
v0y = initial velocity in the y direction (v0 * sin(θ))
ay = acceleration due to gravity (-9.8 m/s^2)
Δy = change in vertical position (h)

Plugging in the values, we get:

0 = (33.7*sin(24))^2 + 2*(-9.8)*110.8

Simplifying and solving for vfy^2:

vfy^2 = -15698.27023

Taking the square root of both sides:

vfy ≈ -125.3 m/s

Note that the negative sign indicates that the final velocity is directed downward.

Step 2: Find the time (t) it takes to achieve this final vertical velocity.
We can use the vertical motion equation:

vfy = v0y + ay * t

Plugging in the values, we have:

-125.3 = 33.7*sin(24) + (-9.8)*t

Simplifying and solving for t:

-9.8t = -125.3 - 33.7*sin(24)
t ≈ 4.713 seconds

Step 3: Find the horizontal displacement (dx).
To find the horizontal displacement, we can use the horizontal motion equation:

dx = v0x * t

Where:
v0x = initial velocity in the x direction (v0 * cos(θ))
t = time

Plugging in the values, we get:

dx = 33.7*cos(24)*4.713
dx ≈ 140.2 meters

Therefore, the horizontal displacement from the base of the building is approximately 140.2 meters.