A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8 s, a stone is released from the balloon. The stone will, after release from the balloon,
Answers
S = ut + (1/2) at²
S = Distance
u = initial speed = 0
t = time = 8 sec
a = acceleration = 1.25 m/s²
S = ut + (1/2)at²
S = 0 + (1/2)(1.25)×8²
=> S = 32 × 1.25
=> S = 40 m
it Means Stone is released from a height of 40 m
released so u = 0 a = g t = ?
40 = 0 + (1/2)gt²
=> t² = 80/g
g = 9.8 m/s²
=> t² = 80/9.8
=> t² = 8.163
=> t = 2.857 sec
Answer:
Explanation:
Answer:
Explanation:
Velocity after rising height h
v = sqrt(2aS) ………[∵ initial velocity of balloon is zero]
v = sqrt(2 × g/8 × h)
v = sqrt(gh/4)
This is the initial velocity of stone.
Time taken by stone rise extra height and come to it’s point of projection is
t1 = 2u / g
= 2 × sqrt(gh/4) / g
= sqrt(h / g)
Final velocity of stone when it touches the ground
v’ = sqrt(u^2 + 2aS)
= sqrt[(gh/4) + 2gh]
= sqrt[2.25 gh]
= 1.5 × sqrt(gh)
Time taken by stone to fall down from point of projection
t2 = (v’ - u) / a
= [(1.5 × sqrt(gh)) - sqrt(gh/4)] / g
= sqrt(gh) [1.5 - 0.5] / g
= sqrt(h / g)
Total time taken for stone to reach the ground
T = t1 + t2
= 2 sqrt(h / g)