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A stone is dropped from the top of a tower 100m in height and at the same time another stone is projected vertically upwards from the ground with a velocity of 25ms find where and when the two stone?
Wiki User
∙ 14y ago
The position of an object in a gravity field is given by: S = 1/2 * a *t ^2 + v * t + x So making the assumption that both are subject to Earth's gravity, a = - 9.81 m/s^2 (negative acting down). Assuming the top stone starts at rest, V_top = 0 m/s, V_bottom = 25 m/s. Taking the problem statement, x_top = 100 m and x_bottom = 0 m. If you want the time when they impact one another, set the two position equations equal and solve for t: 1/2* - 9.81 * t ^2 + 0 + 100 = 1/2 * -9.81 * t^2 + 25 * t + 0 So the acceleration terms cancel, and t = 4 seconds. To check, solve both position equations at t = 4 s. S = 1/2 * -9.81 * 4 ^2 + 0 + 100 = 21.52 m S = 1/2 * -9.81 * 4 ^ 2 + 25 * 4 + 0 = 21.52 m So there you go. The time is 4 seconds, the position is 21.52 m above the ground.
Wiki User
∙ 14y ago
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