Question

a ball of mass 0 .5 kg is dropped from the height of 10 m . the height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is___m . [use g=10m²/s]

Answer 1

The height at which the velocity of the ball equals the acceleration due to gravity (10 m/s^2) is 5 meters, derived using the kinematic equation v^2 = u^2 + 2gh for an object in free fall.

Step-by-step explanation:

To determine the height at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity (g), we use the following kinematic equation for an object in free fall:

v^2 = u^2 + 2gh

Where:

• v is the final velocity
• u is the initial velocity (which is 0 m/s since the ball is dropped)
• g is the acceleration due to gravity (10 m/s^2 in this case)
• h is the height fallen

Since we are looking for the point where the velocity magnitude equals g, we set v = g, or v = 10 m/s:

10^2 = 0^2 + 2(10)h

100 = 20h

h = 5 m

Therefore, the ball will have a velocity equal to the gravitational acceleration after falling from a height of 5 m, which is halfway down from the initial height of 10 m.