a stone is thrown stright upward with speed of 20m/s. it is caught on its way down at a point 15 m above where it was released where it was thrown, how fast was it going when it was caught
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so by we can calculate height
v^2 = u^2+2as but here g is - 10 and v^2 is = 0 so u r answer is 400- 2*10*s so u get as s = 20 . then as per given it caught at height 15m from it throw so. here u^2 is 0 then v ^2 = 2*10*20 which is equal to 20m/s
v^2 = u^2+2as but here g is - 10 and v^2 is = 0 so u r answer is 400- 2*10*s so u get as s = 20 . then as per given it caught at height 15m from it throw so. here u^2 is 0 then v ^2 = 2*10*20 which is equal to 20m/s
SIDDH7456:
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but its asnwer is 10 m/s
how is it possible isn't my steps are correct
it seem correct but i have its answers which is 10 m/s and i dont know how
I don't know hope answer is correct u can ask me also questions of math physics and chemistry
i have poste some question can u plz solve it as well
yeah always
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Answer:
Hey mates
Here is your answer
so by we can calculate height
v^2 = u^2+2as but here g is - 10 and v^2 is = 0 so u r answer is 400- 2*10*s so u get as s = 20 . then as per given it caught at height 15m from it throw so. here u^2 is 0 then v ^2 = 2*10*20 which is equal to 20m/s.
I hope this answer help you
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