An accident occurred on the tollway. A driver accidentally passed through a faulty barricade on a bridge (quite, unfortunately) and landed in a pile of hay (quite, fortunately). Measurements at the accident scene revealed that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. How fast, in miles per hour, was the car going when it went through the barricade. Assume that the contact with the barricade did not slow the car down. (1.00 m/s = 2.24 mi/hr) (Hint: You need time to solve this problem.)

72.53 mi/hr

Explanation:

From the question given above, the following data were obtained:

Vertical distance i.e Height (h) = 8.26 m

Horizontal distance (s) = 42.1 m

Horizontal velocity (u) =?

Next, we shall determine the time taken for the car to get to the ground.

This can be obtained as follow:

Height (h) = 8.26 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

8.26 = ½ × 9.8 × t²

8.26 = 4.9 × t²

Divide both side by 4.9

t² = 8.26 / 4.9

Take the square root of both side by

t = √(8.26 / 4.9)

t = 1.3 s

Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:

Horizontal distance (s) = 42.1 m

Time (t) = 1.3 s

Horizontal velocity (u) =?

s = ut

42.1 = u × 1.3

Divide both side by 1.3

u = 42.1 / 1.3

u = 32.38 m/s

Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:

1 m/s = 2.24 mi/hr

Therefore,

32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s

32.38 m/s = 72.53 mi/hr

Thus, the car was moving at a speed of

72.53 mi/hr.

answer; (d) spontaneous processes are characterized by the overall conversion of order to disorder;

double the distance = divide strength by 4