can anyone plzzz answer
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1
150 bz at t=5or6
distance remain 75,same in both that means at this time period parcle comes to its maximum height i.e 75
by putting in
ok
put one by one value of t=1,2,3,so on in eq.y:20t-t^2
as y is the distance
at t=5andt=6
in both y is same hen above solution
now o come at rest acc. to distance would be double so 2*75=150
distance remain 75,same in both that means at this time period parcle comes to its maximum height i.e 75
by putting in
ok
put one by one value of t=1,2,3,so on in eq.y:20t-t^2
as y is the distance
at t=5andt=6
in both y is same hen above solution
now o come at rest acc. to distance would be double so 2*75=150
shashank4412:
thank you
but the answer is wrong
it will be 100
Answered by
0
y = 20t - t²
v = dy/dt = 20 - 2t
0 = 20 - 2t
t = 10
Particle will stop after 10 seconds
y = 20t - t²
= (20 × 10) - (10)²
= 200 - 100
= 100 m
Distance travelled in 10 seconds is 100 m
(2) is the correct option.
v = dy/dt = 20 - 2t
0 = 20 - 2t
t = 10
Particle will stop after 10 seconds
y = 20t - t²
= (20 × 10) - (10)²
= 200 - 100
= 100 m
Distance travelled in 10 seconds is 100 m
(2) is the correct option.
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