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.