Physics, asked by harinamsingh2919, 11 hours ago

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

Answered by 0496sid
0

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

__________________________________________________________

                               NOT RELATED TO THE ANSWER

  When displacement (in first 3sec) is asked the answer will be 20-5=15m

__________________________________________________________

Answered by sekarvij987
0

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

Similar questions