A body projected vertically up with a speed of 20m/s . The distance travelled by it in 3 sec is (g=10)
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Answered by
3
Answer:
Explanation:
Time of ascent = u/g
= 20/10
= 2 sec
Maximum height= u²/2g
= 20²/2*10
= 20m
During fall, distance covered in one second,
= 1/2 * 10 * 1²
= 5m
Total distance covered in 3 second
= 20+5
=25 m
Answered by
1
Answer:
Explanation:
Time of ascent = u/g
= 20/10
= 2 sec
Maximum height= u²/2g
= 20²/2*10
= 20m
During fall, distance covered in one second,
= 1/2 * 10 * 1²
= 5m
Total distance covered in 3 second
= 20+5
=25 m
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