A tornado lifts a car to a height of 125m above the ground. The tornado flings the car horizontally with a speed of 90.0 m/s. How long does the car take to reach the ground?
To find the time it takes for the car to reach the ground, we need to consider the height from which the car is dropped and the acceleration due to gravity.
1. Determine the initial vertical velocity of the car:
Since the car is lifted to a height of 125m, we can assume it has an initial vertical velocity of 0 m/s because it is at rest.
2. Calculate the time it takes for the car to fall to the ground:
Using the equation for vertical free fall motion,
s = ut + (1/2)gt^2
where:
s = distance (height)
u = initial vertical velocity (0 m/s in this case)
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values,
125 = 0t + (1/2)(9.8)(t^2)
Simplifying the equation,
4.9t^2 = 125
Rearranging the equation,
t^2 = 125 / 4.9
Calculating t,
t = √(125 / 4.9)
t ≈ 5.05 s
Therefore, it will take approximately 5.05 seconds for the car to reach the ground.
To determine how long the car takes to reach the ground, we can use the equations of motion.
Let's consider the vertical motion of the car. The key information we have is that the car is lifted to a height of 125m. We'll assume there is no initial vertical velocity, and the only force acting on the car is gravity.
Using the equation for displacement in vertical motion:
h = ut + (1/2)gt^2
where:
h = vertical displacement (125m)
u = initial vertical velocity (0m/s)
g = acceleration due to gravity (-9.8m/s^2)
t = time
Plugging in the values, the equation becomes:
125 = 0 + (1/2)(-9.8)t^2
Simplifying further, we have:
125 = -4.9t^2
Now, let's solve for t:
-4.9t^2 = 125
Dividing both sides by -4.9, we get:
t^2 = -125 / -4.9
t^2 = 25.51
Taking the square root of both sides, we find:
t = √25.51
t ≈ 5.05 seconds
So, it takes approximately 5.05 seconds for the car to reach the ground.