a balloon is moving upward with a velocity of 3m/s and a constant acceleration of 1.5m/s^2. object of mass 0.25 kg has been dropped from the balloon. find the distance covered by the object at the end of 5th second.
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Hey..Here is your answer u r looking for..
u =3m/s
a =1.5m/s^2
t = 5s
s =ut +1/2at^2
s= 3×5+1/2×1.5×5^2
s=15+0.75×25
s=15+18.75
s=33.75m
since,object is on the baloon and distance of baloon from the earth's surface .hence object is also 33.75m above the earth's surface..
u =3m/s
a =1.5m/s^2
t = 5s
s =ut +1/2at^2
s= 3×5+1/2×1.5×5^2
s=15+0.75×25
s=15+18.75
s=33.75m
since,object is on the baloon and distance of baloon from the earth's surface .hence object is also 33.75m above the earth's surface..
no it will be done by 4th eqn. of motion
i only read 3 equation
get the 3rd one on net I said it can be when when u like to collect extra knowledge for what u are studying
sorry 4th one
Thanks a lot bro
for selecting my answer as brainliest
but solution is wrong
now what happen
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