A small steel ball falls from rest through a distance of 3m. When calculating the time of fall, air resistance can be ignored because

The correct answer is (d) the weight of steel ball is much larger than air resistance.

since the density of steel ball is quite higher than that of air. The weight of even a small steel ball will be much larger than the air resistance acting opposite to the motion of the steel ball. Hence in the case of a freely falling steel ball the air resistance can be neglected.

Learn more about air resistance:

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First, let's write the movement equations for this ball.

The only force acting on the ball will be the gravitational acceleration (because we ignore the air resistance) then the acceleration equation is:

a(t) = -9.8m/s^2

Where the minus sign is because the ball is falling down.

To get the velocity of the ball, we need to integrate over time to get:

v(t) = -(9.8m/s^2)*t + v0

Where v0 is the initial velocity of the ball. Because it falls from rest, we can conclude that the initial velocity is 0 m/s, then the velocity equation is:

v(t) = -(9.8m/s^2)*t

For the position equation we need to integrate again, here we get:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + p0

Where p0 is the initial position. In this cse we know that the ball falls from a height of 3m, then po = 3m

The position equation is:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m

The ball will hit the ground when p(t) = 0m, then we need to solve the equation:

p(t) = -(1/2)*(9.8m/s^2)*t^2 + 3m = 0m

for t.

-(1/2)*(9.8m/s^2)*t^2 + 3m = 0m

3m = (1/2)*(9.8m/s^2)*t^2

3m*2 = (9.8m/s^2)*t^2

6m/(9.8m/s^2) = t^2

√(6m/(9.8m/s^2)) = t = 0.78s

The ball needs 0.78 seconds to hit the ground.

i was thinking more like reflects green light, just not sure

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