You throw a ball of mass 1 kilogram upward with a velocity of a=25 m/s on mars, where the force of gravity is g=3.711 m/s2. Use your calculator to approximate how much longer the ball is in the air on mars.
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consider Newton equation of motion
v=u + at
here v is the velocity at maximum height which will be zero, V = 0
so, 0= u + at
t = -u/a
since a = - g
therefore, t = u/g
on Earth t = u/g = 25/9.8 = 2.551 sec
on Mars t = u/g = 25/3.711 = 6.736sec
now, (t on mars) - (t on earth)
= 6.736-2.551 = 4.185 sec
therefore ball remains 4.185 seconds longer on mars
v=u + at
here v is the velocity at maximum height which will be zero, V = 0
so, 0= u + at
t = -u/a
since a = - g
therefore, t = u/g
on Earth t = u/g = 25/9.8 = 2.551 sec
on Mars t = u/g = 25/3.711 = 6.736sec
now, (t on mars) - (t on earth)
= 6.736-2.551 = 4.185 sec
therefore ball remains 4.185 seconds longer on mars
hrishikeshdeuri:
hi
yes?
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