A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in . If it is projected vertically downwards from the same point with the same speed, it reaches the ground in . Time required to reach the ground, if it is dropped from the top of the tower, is :

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We have a tower of height H. A body is projected vertically upward with speed v zero, taking time t one to reach the ground. When projected downward with the same speed, it takes time t two. We need to find the time t when the body is simply dropped. We use the kinematic equation s equals u t plus half g t squared. Taking downward as positive, the displacement to ground is H. For a dropped body, initial velocity u is zero, giving H equals half g t squared. For downward projection with speed v zero, H equals v zero t two plus half g t two squared. For upward projection, initial velocity is negative v zero, so H equals negative v zero t one plus half g t one squared. For upward projection, the motion has three phases. First, the body goes up taking time v zero over g. Then it falls back to tower level, taking the same time. Finally, it falls from tower to ground, which is exactly the same as the downward projection case, taking time t two. Therefore, total time t one equals two v zero over g plus t two. Solving this gives v zero equals g times t one minus t two, all divided by two. Now we substitute our expression for v zero into the downward projection equation. Starting with H equals v zero t two plus half g t two squared, we replace v zero with g times t one minus t two, all over two. This gives us H equals half g times t one minus t two times t two, plus half g t two squared. Factoring out half g and expanding the brackets, we get H equals half g times t one t two minus t two squared plus t two squared. The t two squared terms cancel out, leaving us with H equals half g t one t two. Now we have two expressions for the height H. From the dropped case, H equals half g t squared. From our analysis of the projections, H equals half g t one t two. Setting these equal gives us half g t squared equals half g t one t two. Canceling half g from both sides, we get t squared equals t one t two. Taking the square root of both sides gives us our final answer: t equals the square root of t one t two. This means the time to reach the ground when dropped is the geometric mean of the times for upward and downward projections.