# Motion under Gravity

Motion under gravity refers to the movement of an object whose vertical motion is affected by the presence of gravity. The force that attracts objects downwards is GRAVITY. In fact, gravity works towards the centre of the Earth. The mass of the object and the gravitational constant determines the magnitude of the gravitational force. Recall that

**WEIGHT = MASS multiplied by GRAVITY.**

Mathematically speaking if weight=W, mass=m and gravity=g then:

**W=mg.**

Weight is a force and is measured in Newtons or kg m/s/s – see Forces & Newton’s First Law. Mass is measured in kilograms (kg) and gravity is an acceleration (m/s/s). For objects reasonably close to the Earth’s surface, gravity is approximated as 9.8 m/s/s (in some questions it may say use g=10 m/s/s). In the absence of air resistance, like in a vacuum, all objects will fall to the ground at the same rate regardless of their mass, shape and size.

Note that weight is different on other planets or celestial objects. Consider the moon, for example. The moon is much smaller in size than the Earth and the gravitational constant is approximately 1.62 m/s/s. As a result, objects weight roughly 17% of what they do on Earth (see more on Gravitation of the Moon). This is why you often see footage of astronauts slowly bouncing over the moon’s surface – gravity is less forceful and the astronauts are lighter there.

**Unlike weight, an object’s mass is always the same.**

## An example of Motion under Gravity

Consider a ball that rolls off the edge of a table and falls for 0.5 seconds before it hits the floor. We can find the velocity of the ball when it hits the ground. The ball starts with zero velocity i.e. 0 m/s in the vertical direction. Assuming that gravity is constant and there is no air resistance, the ball will accelerate downwards at a rate of 9.8 m/s/s. If the ball accelerates this way for 0.5 seconds then the final downward speed of the ball will be

$9.8\times 0.5=4.9$ m/s

Note that this is essentially use of the first SUVAT equation. If gravity is constant, any of the SUVAT equations can be used to solve for motion under gravity.

## Motion under Gravity Examples

Calculate the height of the table in the above example. How tall would the table have to be for the ball to have a speed of 7 m/s when it hits the floor and for how long will it fall?

Use SUVAT equation 2: S=(U+V)T/2. The initial velocity U=0, final velocity V=4.9, time T=1.5 and so the original height is the same as the displacement:

$ S=(0+4.9)\times 0.5/2 = 1.225$

Hence, the original height of the table is 1.225 metres.

We can use SUVAT equation 1(V=U+AT) and SUVAT equation 2(S=(U+V)T/2) to calculate the time taken and the height of the table respectively if the ball reaches 7m/s:

$ 7=0+9.8T \hspace{5pt}\Longrightarrow \hspace{7pt} T=7/9.8=0.714$ to 3 d.p.

$ S=(0+7)\times 0.714/2 = 2.5$

Ultimately, in order to reach 7m/s, the ball must fall 2.5 metres; this takes a total of 0.714 seconds to 3 decimal places.

**Note 1**: An object with motion under gravity does not have a final velocity of 0 when it hits another object.

**Note 2**: Choose whether up is positive or negative and stick to it. If up is positive, gravitational acceleration will be g=-9.8.

An object is propelled upwards with an initial speed of 100km/hr from a point O.

- Find the maximum height reached above the point O.
- The times when the object is 5 metres above the point O.

- The speed has to be converted to m/s first. Namely, 100 kilometres every hour is the same as 100,000 metres every 3,600 seconds or 27.778 m/s to 3 d.p. If we measure displacement upwards, acceleration is -9.8 m/s/s. The height above O can be obtained from SUVAT equation 3:$V^2=U^2+2AS \hspace{3pt}\longrightarrow\hspace{3pt} 0^2=27.778^2+2\times-9.8\times S$
$\hspace{3pt}\Longrightarrow\hspace{3pt}S=\frac{-27.778^2}{-2\times 9.8}=39.368$

As a result, the object reaches a maximum height of 39.368 metres (to 3 d.p.) above the point O.

- The times when the object is 5 metres above the point O can be found using SUVAT equation 4. $S=UT+\frac{1}{2}AT^2\hspace{3pt}\longrightarrow\hspace{3pt}5=27.778\times T+\frac{1}{2}\times-9.8\times T^2$
$\Longrightarrow 4.9T^2-27.778T+5=0$

This can be solved using the quadratic formula to get T=0.186, 5.483.

Consequently, the object is 5 metres above the point O 0.186 seconds (on its way up) and 5.483 seconds (on its way back down) after it is projected to 3 d.p.