Question

A stone is dropped from the top of a building and at the same time another stone is thrown vertically upwards from the ground level with a velocity of 20m/s. The stone meet at exactly 5 seconds after their release. Calculate the height of the building taking g to be 9.8m/s²

Answer 1

The height of the building is 50 meters.

Step-by-step explanation:

In order to solve this problem, we can use the equation of motion for both stones. For the stone that is dropped, we can use the equation:

h = 0.5 * g * t^2

Where h is the height of the building, g is the acceleration due to gravity, and t is the time taken.

For the stone that is thrown vertically upwards, we can use the equation:

h = v0 * t - 0.5 * g * t^2

Where v0 is the initial velocity of the stone.

Since the stones meet at exactly 5 seconds after their release, we can set the equations equal to each other and solve for h:

0.5 * g * (5)^2 = 20 * 5 - 0.5 * g * (5)^2

Simplifying the equation gives:

0.5 * g * (5)^2 = 20 * 5 - 0.5 * g * (5)^2

5 * 5 * g = 20 * 5

g = 20 * 5 / (5 * 5)

g = 20 / 5

g = 4 m/s^2

Thus, the height of the building is given by:

h = 0.5 * g * (5)^2

h = 0.5 * 4 * (5)^2

h = 0.5 * 4 * 25

h = 50 meters