A ball is thrown up with a speed of 20m / s from the ground. Find distance traveled after 3 seconds (Take g = 10m / s²)
Answers
Answer:
25m
Explanation:
Distance travelled during ascent 'h'
v²-u² = 2gh
0-20² = 2×(-10)×h
h = 20m
Time taken for ascent 't'
t = v-u/a = 0-20/(-10) [∵ Acceleration during ascent = -g]
t = 2sec
Distance travelled during first 3 seconds
= h + distance travelled in the next 1 second (or) d in 1st second of descent
=20 + [(0)(1)+(1/2)(10)(1²)] [∵ s=ut+1/2gt², a=g during descent]
=25m
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NOT RELATED TO THE ANSWER
When displacement (in first 3sec) is asked the answer will be 20-5=15m
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Answer:
answer is : 60m
Explanation:
because the ball thrown from the ground as speed 20m/s . In 3s so 20m×3 the final answer is 60m