A stone is released from top of a tower. If it
covers a distance of 80 m in last 2 sec of
it's motion, then the height of the tower is:
(1) 320 m
2)125 m
(3) 180 m
(4) 245 m
Answers
Answer:
This question is super easy if you’ve been taught kinematics… if you know equations of motion. Since you’re asking this, I'll assume you haven't been taught kinematics. So… how to figure this out from scratch? (We’ll still need to know some basic algebra.)
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The split second that we first let go of the stone, it’ll have no speed at all. So its initial speed, let’s call it u, is zero, i.e. u = 0 m/s.
As the stone is falling, it’ll speed up due to gravity. Let's call that acceleration g. On the Earth’s surface (i.e. at sea level) and even 90 m above it, it'll be roughly true that the free falling stone will speed up by 9.81 m/s each second. (You can look up this value online.) In short, its acceleration is roughly a steady, unchanging:
g = 9.81 m/s every second.
In other words, after 1 second, the stone’s final speed, v, would be v =(9.81 × 1) m/s.
After 2 seconds, the stone’s speed would be v = (9.81 × 2) m/s.
After 3 seconds, the stone’s speed would be v = (9.81 × 3) m/s.
In general, after t seconds, the stone’s speed would be v = 9.81 × t = gt.
Since v = gt, then t = v/g. Let’s call this equation (1).
Let’s call the distance travelled by the stone x. If the stone was moving at a constant speed, i.e. if v was constant and g was 0, then:
After 1 second, the stone would have travelled a distance of x = (v × 1) m.
After 2 seconds, the stone would have travelled x = (v × 2) m.
After 3 seconds, the stone would have travelled x = (v × 3 )m.
In general, after t seconds, the stone would have travelled x = vt.
But we know from earlier that the stone keeps speeding up as it falls, i.e. v is not constant. So we could instead look at the average speed, let’s call it <v>, and say:
After t seconds, the stone would have travelled a distance of x = <v> × t. Let’s call this equation (2).
Combining (1) and (2): x = <v> × v/g. Let’s call this equation (3).
We know by the end, the stone travels a distance of x = 90 m. We know that it speeds up by g = 9.81 m/s each second that it falls. We want to find its final speed, v, just before it hits the ground. The only missing piece of the puzzle in (3) is the stone’s average speed, <v>.
If we had several boxes with a different number of items in each box, to find the average number of items per box, we’d add up all the items and divide the total by the number of boxes. E.g.: If we had 2 boxes with 5 and 3 items each, the average would be (5 + 3)/2 = 4 items per box.
Since the acceleration is constant, we can use the same method as the items in the boxes to find the average speed, <v>. We just take the initial speed, u, and the final speed, v, add them up, and divide them by 2 to get:
<v> = (u + v)/2. But u = 0. So, <v> = v/2. Let’s call this equation (4).
(Note: If the acceleration wasn’t constant, finding the average speed would be much trickier. We wouldn’t be able to use the easy method above.)
Combining (3) and (4): x = (v/2) × (v/g) = (v^2)/2g
So (v^2) = 2gx and v = √(2gx)
So the stone’s final speed just before hitting the ground is:
v = √(2 × 9.81 × 90) m/s = √(1765.8) m/s = 42.0 m/s.