A balloon starts rising from ground from rest with an upward acceleration 2 m/s2. Just after 1 second, a stone is dropped from it. The time taken by stone to strike the ground is nearly?
Answers
Answered by
7
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
Initial velocity of balloon(u) = 0
Resultant acceleration of balloon(a) = +2 (m/s)/s
Height of balloon after t (= 1 second) = (1/2) at^2=2*1/2 = +1m
Velocity of balloon at that time = 0 + [+2(m/s)/s](1s) = +2 m/s
Initial acceleration when stone is left = -9.8 m/s
Initial velocity of stone = final velocity after given time = +2 m/s
Height travelled by balloon = -(height reached by balloon after given time) = -1 m
But,
by using s = ut + (1/2) at^2
(I remove units to simplify answer)
-1 = 2t + (1/2)(-9.8) t^2
4.9t^2 -2t -1 = 0
t = [2 ± √{4 + (4*4.9)}]/9.8 = [1 ± √(1 + 4.9)]/4.9 = [1 ± √5.9]/4.9
But, t is always +ve
t=(1+ √5.9)/4.9
= 0.699794196 ≈ 0.70s
So, time taken = 0.699794196 ≈ 0.70s
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
Initial velocity of balloon(u) = 0
Resultant acceleration of balloon(a) = +2 (m/s)/s
Height of balloon after t (= 1 second) = (1/2) at^2=2*1/2 = +1m
Velocity of balloon at that time = 0 + [+2(m/s)/s](1s) = +2 m/s
Initial acceleration when stone is left = -9.8 m/s
Initial velocity of stone = final velocity after given time = +2 m/s
Height travelled by balloon = -(height reached by balloon after given time) = -1 m
But,
by using s = ut + (1/2) at^2
(I remove units to simplify answer)
-1 = 2t + (1/2)(-9.8) t^2
4.9t^2 -2t -1 = 0
t = [2 ± √{4 + (4*4.9)}]/9.8 = [1 ± √(1 + 4.9)]/4.9 = [1 ± √5.9]/4.9
But, t is always +ve
t=(1+ √5.9)/4.9
= 0.699794196 ≈ 0.70s
So, time taken = 0.699794196 ≈ 0.70s
✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️✳️
Similar questions