What is the final velocity of a ball dropped from a height of 50 m just before it hits the ground?

To determine the final velocity of a ball dropped from a height of 50 meters just before it hits the ground, we can use the principle of conservation of energy or the equations of motion under gravity.

When an object is dropped, it is subjected to the acceleration due to gravity, which is approximately 9.81 m/s². Assuming there is no air resistance acting on the ball, we can use the formula derived from the equations of motion:

[ v^2 = u^2 + 2as ]

Where:

• ( v ) is the final velocity,

• ( u ) is the initial velocity (which is 0 m/s since the ball is dropped),

• ( a ) is the acceleration (9.81 m/s²),

• ( s ) is the distance fallen (50 m).

Plugging in the values, we get:

[ v^2 = 0^2 + 2 \times 9.81 , \text{m/s}^2 \times 50 , \text{m} ]

[ v^2 = 2 \times 9.81 \times 50 ]

[ v^2 = 981 ]

Taking the square root of