a small plane takes off at a constant velocity of 50 km/h of 37 degrees. at 3.00 s (a) how high is the plane above the groundand (b)what horizontal distance has the plane travelad from the liftoff point?
Find the vertical components, and the horizontal components (using sin, cosine functions),
Then distance=velocity*time
To find the height and horizontal distance traveled by the plane, we can use the principles of trigonometry and kinematics.
(a) To determine how high the plane is above the ground at 3.00 seconds, we need to calculate the vertical component of its velocity. Since the plane took off at a constant velocity, we can assume its acceleration is zero.
We can use the following kinematic equation to find the vertical displacement (height):
y = v0y * t + (1/2) * a * t^2
where:
y = vertical displacement (height)
v0y = initial vertical component of velocity = v0 * sin(theta)
t = time = 3.00 s
a = acceleration = 0 (since the plane's velocity is constant)
theta = angle of takeoff = 37 degrees
v0 = initial velocity = 50 km/h
First, let's convert the initial velocity to m/s:
50 km/h = 50 * (1000 m / 1 km) / (3600 s / 1 h) ≈ 13.9 m/s
Now, calculate the initial vertical component of velocity:
v0y = 13.9 m/s * sin(37 degrees)
Using a calculator:
v0y ≈ 13.9 m/s * 0.6018 ≈ 8.37 m/s
Plugging in the values:
y = 8.37 m/s * 3.00 s + (1/2) * 0 * (3.00 s)^2
y ≈ 25.11 m
Therefore, the plane is approximately 25.11 meters above the ground at 3.00 seconds.
(b) To determine the horizontal distance traveled by the plane from the liftoff point, we need to calculate the horizontal component of its velocity.
We can use the following equation to find the horizontal displacement:
x = v0x * t
where:
x = horizontal displacement
v0x = initial horizontal component of velocity = v0 * cos(theta)
t = time = 3.00 s
theta = angle of takeoff = 37 degrees
v0 = initial velocity = 13.9 m/s
First, calculate the initial horizontal component of velocity:
v0x = 13.9 m/s * cos(37 degrees)
Using a calculator:
v0x ≈ 13.9 m/s * 0.7986 ≈ 11.08 m/s
Now, plug in the values to find the horizontal distance traveled:
x = 11.08 m/s * 3.00 s
x ≈ 33.24 m
Therefore, the plane has traveled approximately 33.24 meters horizontally from the liftoff point.