a stone is dropped from a height of 50 m. calculate (i) time taken to reach the ground. (ii) final velocity just before touching the ground (iii) velocity after 0.5s of its motion.
Answers
Answer:
Explanation: 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).
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