A balloon moves vertically upward with an initial speed of 5 m/s and a constant upward acceleration of 2 m/ $s^2.$ At t = 0, a stone is released from the balloon. Exactly 1 s later, a second stone is released from the same balloon. Neglect air resistance and take g = 10 m/ $s^2.$ What is the separation between the two stones 2 s after the second stone is released?
Options

Question

A balloon moves vertically upward with an initial speed of 5 m/s and a constant upward acceleration of 2 m/ $s^2.$ At t = 0, a stone is released from the balloon. Exactly 1 s later, a second stone is released from the same balloon. Neglect air resistance and take g = 10 m/ $s^2.$ What is the separation between the two stones 2 s after the second stone is released?

Options

G1 Physics · Accepted Answer

Take the position of the balloon at t = 0 as y = 0 and upward as positive. First stone is released with u1 = 5 m/s, so at t = 3 s: y1 = 5(3) - 5(3)^2 = -30 m. At t = 1 s, the balloon is at y = 5(1) + (1/2)(2)(1)^2 = 6 m and has speed 5 + 2(1) = 7 m/s. The second stone is released there with u2 = 7 m/s. Two seconds later: y2 = 6 + 7(2) - 5(2)^2 = 0. Separation = 0 - (-30) = 30 m. Hence option C is correct.