How to Find Velocity of a Falling Object

Close to the Earth’s surface, a falling object experiences a constant downward acceleration g of approximately 9.81 ms-2. If we assume air resistance to be negligible, we can use the equations of motion for an object experiencing a constant acceleration to analyse the kinematics of the particle. Furthermore, to make matters simple, we will assume that the particle is moving along a line.

When doing typical calculations of this type, it is important to define a direction to be positive. Then, all vector quantities which point along this direction should be taken to be positive while quantities that point in the opposite direction should be taken to be negative.

How to Find Velocity of a Falling Object, which Started from Rest

For this case, we have v_i = 0. Then, our four equations of motion become:

\Delta x = \frac{1}{2}v(t)t

v(t) = at

\Delta x = \frac{1}{2}gt^2

v(t)^2 = 2a\Delta x

Example

A stone is dropped from the Sydney Harbour Bridge, which is 49 m above the surface of water. Find the velocity of the stone as it hits the water.

At the beginning, the stone’s velocity is 0. Taking the downwards direction to be positive, we have \Delta x = 49 m and a = g = 9.81 m s-2. Using the fourth equation above, then, we have: v(t) = \sqrt{2\times 9.81\times 49} = 31 m s-1.

How to Find Velocity of a Falling Object, which Did not Start from Rest

Here, the equations of motion apply as usual.

Example

A stone is thrown downward at a speed of 4.0 m s-1 from the top of a 5 m building. Calculate the speed of the stone as it hits the ground.

Here, we use the equation v(t)^2 = {v_i}^2+2a\Delta x. Then, v(t) = \sqrt{{v_i}^2+2a\Delta x}. If we take downwards direction to be positive, then we have v_i = 4.0 m s-1. and a = g = 9.81 m s-2. Substituting the values, we get: v(t) = \sqrt{4.0^2+2 \times 9.81 \times5} = 11 m s-1.

Example

A stone is thrown upward at a speed of 4.0 m s-1 from the top of a 5 m building. Calculate the speed of the stone as it hits the ground.

Here, the quantities are the same as those in the previous example. The displacement of the body is still 5 m s-1 downwards, as the initial and final positions of the stone are the same as those in the earlier example. The only difference here is that the initial velocity of the stone is upward. If we take downward direction to be positive, then we would have v_i = -4 m s-1. However, for this particular case, since v(t) = \sqrt{{v_i}^2+2a\Delta x}, the answer should be the same as before, because squaring -v_i gives the same result as squaring v_i.

Example

A ball is thrown upwards at a speed of 5.3 m s-1. Find the velocity of the ball 0.10 s after it was thrown.

Here, we’ll take upwards direction to be positive. Then, v_i = 5.3 m s-1. The acceleration  g is downward, so a = g = -9.81 m s-2 and time t = 0.10 s. Taking the equation v(t) = v_i+at, we have v(t) = 5.3-9.81 \times 0.1 = 4.3 m s-1. Since we get a positive answer, this means that the ball is still travelling upward.

Let’s now try to find the velocity of the ball 0.70 s after it was thrown. Now, we have: v(t) = 5.3-9.81 \times 0.7 = -1.6 m s-1. Note that the answer is negative. This means that the ball has reached the top, and is now moving down.

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