2.13 Free Fall Acceleration

2.13 Free Fall Acceleration

When a body is falling freely under the action of gravity of the Earth, the acceleration acting on it is the gravitational acceleration and is denoted by g. The direction of gravitational acceleration is always downwards. Its value is 9.8 m s⁻², but for convenience we shall use the value of g as 10 m s⁻².

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For Your Information! Light and heavier objects, when falling through a vacuum, move side by side.

Since freely falling bodies move vertically downwards in a straight line with uniform acceleration g, the three equations of motion can be applied to the motion of such bodies. While applying the equations of motion, the acceleration a is replaced by g. Thus, the equations of motion for freely falling bodies can be written as:

v_f = v_i + gt              ............ (1)
S = v_i t + 1/2 gt²         ............ (2)
2gS = v_f² − v_i²           ............ (3)

It should be remembered that while using these equations, the following points should be kept in mind:

1. If a body is released from some height to fall freely, its initial velocity v_i will be taken as zero.
2. The gravitational acceleration g will be taken as positive in the downward direction. All other quantities will also be taken as positive in the downward direction. The quantities in the direction opposite to the acceleration will be taken as negative.
3. If a body is thrown vertically upward, the value of g will be negative and the final velocity will be zero at the highest point.

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For Your Information! Acceleration of free fall g is 10 m s⁻² for all objects.

Example 2.5

An iron bob is dropped from the top of a tower. It reaches the ground in 4 seconds. Find: (a) the height of the tower, and (b) the velocity of the ball as it strikes the ground.

Solution:

For a freely falling body:

• Initial velocity = v_i = 0
• Acceleration = g = 10 m s⁻²
• Time = t = 4 s
• Height (distance) = S = h = ?
• Final velocity = v_f = ?

(a) According to the second equation of motion:

S = v_i t + 1/2 gt²
h = 0 × 4 s + 1/2 × 10 m s⁻² × (4)² s²
h = 80 m

(b) From the first equation of motion:

v_f = v_i + gt
v_f = 0 + 10 m s⁻² × 4 s
v_f = 40 m s⁻¹

Example 2.6

An arrow is thrown vertically upward with the help of a bow. The velocity of the arrow when it leaves the bow is 30 m s⁻¹. Determine the time to reach the highest point. Also, find the maximum height attained by the arrow.

Solution:

Here, acceleration will be taken as negative because the arrow is thrown vertically upward.

• Initial velocity = v_i = 30 m s⁻¹
• Final velocity = v_f = 0
• Acceleration = g = −10 m s⁻²
• Time = t = ?
• Height = S = h = ?

From the first equation of motion:

v_f = v_i + gt
t = (v_f − v_i) / (−g)
t = (0 − 30 m s⁻¹) / (−10 m s⁻²)
t = 3 s

From the third equation of motion:

2gS = v_f² − v_i²
S = (v_f² − v_i²) / 2(−g)
h = (0 − (30)² m² s⁻²) / (2 × 10 m s⁻²)
h = 45 m