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This year is 60 years since I learned this stuff, and one of the things I always remembered is the formula for the distance a dropped object falls:
D = 1/2 A T²
Distance = (1/2) (acceleration) (time²)
The reason I never forgot it is because it's SO useful SO often. You really should memorize it. And don't bury it too deep in your toolbox ... you'll be needing it again very soon. (In fact, if you had learned it the first time you saw it, you could have solved this problem on your own today.)
The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth. Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .
In 5 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (5 sec)²
D = (4.9 m/s²) (25 sec²)
D = 122.5 meters
In 6 seconds:
D = 1/2 A T²
D = (1/2) (9.8 m/s²) (6 sec)²
D = (4.9 m/s²) (36 sec²)
D = 176 meters
Answer:
18.9 m.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 70 km/h
Height (h) =?
Next, we shall convert 70 km/h to m/s. This can be obtained as follow:
3.6 km/h = 1 m/s
Therefore,
70 km/h = 70 km/h × 1 m/s / 3.6 km/h
70 km/h = 19.44 m/s
Finally, we shall determine the height. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 19.44 m/s
Acceleration due to gravity (g) = 10 m/s²
Height (h) =?
v² = u² + 2gh
19.44² = 0² + (2 × 10 × h)
377.9136 = 0 + 20h
377.9136 = 20h
Divide both side by 20
h = 377.9136 / 20
h = 18.9 m
Thus, the car will fall from a height of 18.9 m
