A balloon starts from the ground and moves vertically upwards with an acceleration of 2
m/s2
. After 5s, a stone is released from the balloon. Find the time taken by the stone to reach the
ground
A balloon starts from the ground and moves vertically upwards with an acceleration of 2
m/s2
. After 5s, a stone is released from the balloon. Find the time taken by the stone to reach the
ground
Here upward acceleration of balloon is 2 m/s² upwards.
u=0.
v=u+at.
So velocity gained by the stone after 5 seconds is 2* 5= 10m/s
v=10m/s.
S=ut+(1/2)at².
Height reached by stone upto the time at which it is released is (1/2) *2*(5²)=25 metre (using equation of motion s=ut+1/2 at^2 here intial velocity u=0.)
Let us suppose the stone takes time T to fall back then using above equation of motion.
v=ú. (initial velocity of stone when dropped).
S=úT + (1/2)gT²
25=-10*T+(1/2)*10*T²
(taking g=10m/s^2 and intial velocity is upward while net displacement is downward so it is aken negative)
5T²–10T-25=0 or
T²–2T-5=0
Solving this we get T=3.45seconds and T=-1.45seconds.
neglecting negative ,
Time required for stone to reach ground =3.45s.
Here upward acceleration of balloon is 2 m/s² upwards.
u=0.
v=u+at.
So velocity gained by the stone after 5 seconds is 2* 5= 10m/s
v=10m/s.
S=ut+(1/2)at².
Height reached by stone upto the time at which it is released is (1/2) *2*(5²)=25 metre (using equation of motion s=ut+1/2 at^2 here intial velocity u=0.)
Let us suppose the stone takes time T to fall back then using above equation of motion.
v=ú. (initial velocity of stone when dropped).
S=úT + (1/2)gT²
25=-10*T+(1/2)*10*T²
(taking g=10m/s^2 and intial velocity is upward while net displacement is downward so it is aken negative)
5T²–10T-25=0 or
T²–2T-5=0
Solving this we get T=3.45seconds and T=-1.45seconds.
neglecting negative ,
Time required for stone to reach ground =3.45s.
