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A body is projected vertically up from a foot of the tower at the same time another body is dropped from top of the tower. If the height of the tower is 392m and the velocity of the projectile is 98m/s. Find when and where the two bodies will meet?
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A body is projected vertically up from a foot of the tower at the same...
Problem Statement

A body is projected vertically up from a foot of the tower at the same time another body is dropped from top of the tower. If the height of the tower is 392m and the velocity of the projectile is 98m/s. Find when and where the two bodies will meet?


Understanding the Problem

The problem involves two bodies, one is dropped from a height of 392m and the other is projected upwards from the foot of the tower. The velocity of the projectile is given as 98m/s. We need to find the time and location where the two bodies will meet.


Approach to Solve


  1. Calculate the time taken by the dropped body to reach the ground.

  2. Calculate the distance covered by the projected body during the time taken by the dropped body to reach the ground.

  3. Find the height at which the two bodies will meet.

  4. Find the time at which the two bodies will meet.



Solution


  1. The time taken by the dropped body to reach the ground can be calculated using the formula,

    time = sqrt(2 * height / g)

    where g is the acceleration due to gravity, which is approximately equal to 9.8 m/s^2.

    Hence, time taken by dropped body, t1 = sqrt(2 * 392 / 9.8) = 8 seconds.

  2. During the time taken by the dropped body to reach the ground, the projected body will cover a distance equal to,

    distance = velocity * time

    Hence, distance covered by the projected body, d1 = 98 * 8 = 784m.

  3. Let h be the height at which the two bodies will meet. The time taken by the projected body to reach this height can be calculated using the formula,

    time = velocity / g

    Hence, time taken by the projected body, t2 = 98 / 9.8 = 10 seconds.

  4. During the time taken by the projected body to reach the height where the two bodies meet, the dropped body would have covered a distance equal to,

    distance = 0.5 * g * t^2

    Hence, distance covered by the dropped body, d2 = 0.5 * 9.8 * (10-8)^2 = 19.6m.

  5. Therefore, the height at which the two bodies will meet is

    h = 392 - (d1 + d2) = 392 - (784 + 19.6) = 588.4m.

    The time at which the two bodies will meet is

    t = t1 + (h / velocity) = 8 + (588.4 / 98) = 14 seconds.



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