Question

A tornado lifts a truck 252 m above the ground. As the storm continues, the tornado throws the truck horizontally. It lands 560 m away from where it was picked up.

How fast was the truck traveling horizontally through its flight? Round to the nearest tenth.

Answer 1

The horizontal speed of the truck was approximately 78.1 m/s, calculated using the distance it traveled horizontally (560 m) and the time of its free fall which was deduced from its vertical drop (252 m) under gravity.

Step-by-step explanation:

To calculate how fast the truck was traveling horizontally through its flight after being lifted by a tornado, we need to use the formula for horizontal speed (v), which is distance (d) divided by time (t): v = d/t. However, we are not provided with the time directly. Instead, we can deduce the time using the formula for free fall under gravity to find out how long it takes for the truck to fall from the height of 252 m back to the ground. This time will be the same as the time the truck spent flying horizontally due to the independence of horizontal and vertical motions.

The formula for the time taken (t) to fall from a height (h) under gravity (g) is derived from h = 0.5 * g * t^2, which gives us t = sqrt(2h/g). Using the height (252 m) and the acceleration due to gravity (9.8 m/s^2), we can calculate the time of fall.

Using h = 252 m and g = 9.8 m/s^2:

1. t = sqrt(2 * 252 m / 9.8 m/s^2)
2. t = sqrt(2 * 252 / 9.8)
3. t = sqrt(51.4286)
4. t = 7.17 s (rounded to two decimal places)

Now we know the time, we can work out the horizontal speed:

1. Using d = 560 m:
2. v = d/t
3. v = 560 m / 7.17 s
4. v = 78.10 m/s

Therefore, the truck was traveling horizontally at a speed of approximately 78.1 m/s.

Answer 2

the answer is 78.1 I already checked it