A lift is moving with velocity 10 m/s upwards with an acceleration 5 m/s2 upwards. At the same instant of time a coin is dropped from 7.5 m above the floor of lift, then the distance travelled by the coin as seen from the ground is ( take g = 10 m/s2)
ag=10 m/s2
uc=0
uL=10 m/s
aL=5 m/s2
let the time taken for the coin to drop be t
then, distance travelled by the lift in time t,
s=10 t +2.5 t2
distance travelled by coin = 5 t2
total distance covered by both lift and coin together = 7.5 m
so, 7.5 t2 + 10 t = 7.5
or, 3 t2 + 4 t = 3
solve for t.
uc=0
uL=10 m/s
aL=5 m/s2
let the time taken for the coin to drop be t
then, distance travelled by the lift in time t,
s=10 t +2.5 t2
distance travelled by coin = 5 t2
total distance covered by both lift and coin together = 7.5 m
so, 7.5 t2 + 10 t = 7.5
or, 3 t2 + 4 t = 3
solve for t.