A Volkswagen runs straight off a cliff. The Volkswagen is traveling at a speed of 39.0 m/s when it leaves the road. If the cliff is 16.0 m high, how far horizontally does the car travel before it smashes into the ground below?
Xo = 39.0 m/s.
h = 0.5g*t^2 = 16 m.
4.9t^2 = 16
t^2 = 3.27
Tf = 1.81 s. = Fall time.
Dx = Xo*Tf = 39m/s * 1.81s = 70.5 m.
Well, that's one way to try a new take on a car commercial! Okay, let's do some math while we try not to think about the poor Volkswagen.
To solve this, we can use a little physics: the time of flight and horizontal distance are related by the equation d = v * t, where d is the distance, v is the velocity, and t is the time.
First, let's find the time it takes for the Volkswagen to reach the ground from the cliff. We can use the equation h = 1/2 * g * t^2, where h is the height, g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time of flight.
Rearranging this equation, we get t = sqrt(2h/g). Plugging in the numbers, we have t = sqrt(2 * 16.0 m / 9.8 m/s^2) = 2.03 seconds.
Now let's find the horizontal distance. We already know that the velocity is 39.0 m/s, which means we need to determine how far the car travels in those 2.03 seconds.
Using the equation d = v * t, we have d = 39.0 m/s * 2.03 s = 79.17 meters.
So, horizontally speaking, the Volkswagen travels approximately 79.17 meters before its epic, and hopefully fictional, smash into the ground.
To find the horizontal distance the Volkswagen travels before it hits the ground, we can use the equations of projectile motion. One of the equations states that the horizontal distance (d) is equal to the horizontal velocity (Vx) multiplied by the time of flight (t).
Given:
Initial velocity (Vi) = 39.0 m/s
Height (h) = 16.0 m
First, we need to find the time of flight.
The time it takes for an object to fall from a certain height can be determined using the following equation, which aids in our calculation since the vertical motion is similar to free fall:
h = (1/2) * g * t^2
Where g is the acceleration due to gravity (assumed to be 9.8 m/s^2) and t is the time of flight.
Rearrange the equation to solve for t:
t^2 = (2h) / g
Substitute the given values:
t^2 = (2 * 16.0 m) / 9.8 m/s^2
t^2 ≈ 3.27 s^2
Take the square root of both sides:
t ≈ √3.27 s
t ≈ 1.81 s
Now, to find the horizontal distance, we use the formula mentioned earlier:
d = Vx * t
Since there is no horizontal acceleration, the horizontal velocity (Vx) remains constant throughout the motion. Therefore, Vx = Vi.
Substitute the given value:
d = 39.0 m/s * 1.81 s
d ≈ 70.59 m
Therefore, the Volkswagen travels approximately 70.59 meters horizontally before it smashes into the ground below.
To find the horizontal distance traveled by the car before it hits the ground, we need to use the principle of projectile motion. Here's how you can find the answer step by step:
Step 1: Split the motion into horizontal and vertical components. In this case, the horizontal motion of the car remains unaffected, while the vertical motion is influenced by gravity.
Step 2: Calculate the time it takes for the car to hit the ground using the vertical motion. To do this, we can use the formula for vertical displacement:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy = vertical displacement (16.0 m)
v₀y = initial vertical velocity (0 since it starts from rest)
a = acceleration due to gravity (-9.8 m/s²)
t = time
Solving for t gives:
16.0 = 0 * t + (1/2) * (-9.8) * t²
16.0 = -4.9t²
Dividing by -4.9:
-3.2653 ≈ t²
Taking the square root yields:
t ≈ 1.8063 seconds
Step 3: Determine the horizontal distance traveled by the car using the horizontal motion. The horizontal distance (d) can be calculated using the formula:
d = v₀x * t
Where:
d = horizontal distance
v₀x = initial horizontal velocity (39.0 m/s)
t = time (1.8063 s)
Multiplying the initial horizontal velocity by the time gives:
d = 39.0 * 1.8063
d ≈ 70.66 meters
Therefore, the car travels approximately 70.66 meters horizontally before smashing into the ground below.