A stone is thrown horizontally from the top of a 20m high tower with a...
Answer:Introduction:
In this problem, we need to find the speed of the stone when it is at the same distance from the top as well as the base of the tower. The stone is thrown horizontally from the top of a 20m high tower with an initial velocity of 10m/s and air drag is negligible.
Approach:
We can use the principle of conservation of energy to solve this problem. At the initial point, the kinetic energy of the stone is zero, and the potential energy is mgh, where m is the mass of the stone, g is the acceleration due to gravity, and h is the height of the tower. At any point during the motion of the stone, the total energy of the stone is the sum of its kinetic and potential energy.
Solution:
When the stone reaches a height of 10m, it has lost half of its initial potential energy. At this point, the kinetic energy of the stone is equal to its potential energy. Therefore, we can equate the two to find the velocity of the stone.
Potential energy at height h = mgh
Initial potential energy = mgh = 20m x 9.8m/s^2 x m = 196mJ
When the stone reaches a height of 10m, its potential energy is:
Potential energy at height 10m = mgh' = 10m x 9.8m/s^2 x m = 98mJ
Kinetic energy of the stone at this point = Potential energy at this point
1/2 mv^2 = 98mJ
v^2 = 196mJ/m = 196m/s^2
v = sqrt(196) = 14m/s
Therefore, the speed of the stone when it is at the same distance from the top as well as the base of the tower is 14m/s.