A car launches itself off a 20° ramp at the top of a 25m cliff travelling at 35 m/s. How long will it be in the air for and how far from the base of the cliff will it land?
Vertical:
hfinal=hinitial+ViSin20*t-4.9t^2
hfinal=0, hinitial=25, Vi=35, solve for t.
how far? distance=Vicos20*t
To find the time the car is in the air and the horizontal distance it will travel, we can break down the problem into two components: vertical motion (up and down) and horizontal motion (forward).
First, let's analyze the vertical motion.
1. Determine the vertical velocity of the car at the top of the ramp.
The car's initial vertical velocity can be determined using trigonometry. Since the car launches off a ramp at an angle of 20° with the horizontal, it will have two components of velocity: one in the horizontal direction and one in the vertical direction. We can find the vertical component of the velocity using the formula:
Vertical velocity (v_y) = initial velocity (v) * sin(θ)
In this case, the initial velocity is 35 m/s, and the angle of the ramp is 20°. So we have:
v_y = 35 m/s * sin(20°)
2. Calculate the time the car will be in the air.
To find the time, we need to know how long it takes for the car to reach its highest point and then return to the ground. We can use the equation for vertical motion:
displacement (Δy) = initial velocity (v_y)t + (1/2) * acceleration (a_y) * t^2
At the highest point of the trajectory, the displacement (Δy) would be equal to the height of the cliff (25m). Also, since the car is launched vertically, the initial velocity in the vertical direction (v_y) is 0. The acceleration due to gravity (a_y) is a constant value of -9.8 m/s^2 (negative because it's acting opposite to the positive direction). So our equation becomes:
25m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
Solving this quadratic equation will give us the time (t) it takes for the car to reach the ground.
3. Calculate the horizontal distance traveled.
Next, we can calculate the horizontal distance traveled by the car. This can be determined using the equation for horizontal motion:
displacement (Δx) = horizontal velocity (v_x) * t
The horizontal velocity (v_x) can be found using the formula:
Horizontal velocity (v_x) = initial velocity (v) * cos(θ)
In this case, the initial velocity is 35 m/s, and the angle of the ramp is 20°. So we have:
v_x = 35 m/s * cos(20°)
Now we can find the horizontal displacement (Δx) by multiplying the horizontal velocity by the time of flight (t) obtained in the previous step.
Now, let's go ahead and calculate all these values step by step.
Step 1: Calculate the vertical velocity (v_y):
v_y = 35 m/s * sin(20°)
v_y ≈ 11.96 m/s (rounded to two decimal places)
Step 2: Calculate the time (t) by solving the quadratic equation:
25m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
This equation can be rearranged as:
4.9t^2 = 25m
t^2 ≈ 5.102
t ≈ √5.102 ≈ 2.26 s (rounded to two decimal places)
Step 3: Calculate the horizontal velocity (v_x):
v_x = 35 m/s * cos(20°)
v_x ≈ 32.80 m/s (rounded to two decimal places)
Step 4: Calculate the horizontal displacement (Δx):
Δx = v_x * t
Δx ≈ 32.80 m/s * 2.26 s
Δx ≈ 74.13 m (rounded to two decimal places)
Therefore, the car will be in the air for approximately 2.26 seconds and will land approximately 74.13 meters from the base of the cliff.