Gravitation | Equation of Motion of freely falling Bodies

Equation of Motion of freely falling Bodies

When the bodies are falling under influence of gravity, they experience acceleration g i.e. 9.8 $\displaystyle \mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ . However, when these are going up against gravity they move with retardation of 9.8 $\displaystyle \mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ .

All the gravitation of motion already read by us are valid for freely falling body with the different that a is replaced by g. For motions vertically upwards a is replaced by – g. Here ‘s’ is also replaced by h, the height.

$\displaystyle v=u+at$ changes to $\displaystyle v=u+gt$

$\displaystyle s=ut+\frac{1}{2}a{{t}^{2}}$ changes to $\displaystyle h=ut+\frac{1}{2}g{{t}^{2}}$

$\displaystyle {{v}^{2}}={{u}^{2}}+2as$ changes to $\displaystyle {{v}^{2}}={{u}^{2}}+2gh$

Question: A car falls off a ledge and drops to the ground in 0.5s. Let $\displaystyle g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ .

(i) What is its speed on striking the ground?

(ii) What is its average speed during the 0.5 s?

(iii) How high is the ledge from the ground?

Solutions: Time, $\displaystyle t=\frac{1}{2}$ second

Initial velocity, $\displaystyle u=0\,\,\mathbf{m/s}$

Acceleration due to gravity, $\displaystyle g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Acceleration of the car, $\displaystyle a=+10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ (downward)

(i) Speed $\displaystyle v=u+gt$

$\displaystyle v=0+(10\times 0.5)$

$\displaystyle v=5\,\,\mathbf{m/s}$

(ii) Average speed $\displaystyle =\frac{u+v}{2}$

$\displaystyle =\frac{0+5}{2}$ $\displaystyle =2.5\,\mathbf{m/s}$

(iii) Distance traveled $\displaystyle (S)=ut+\frac{1}{2}a{{t}^{2}}$

$\displaystyle =0\times 0.5+\frac{1}{2}\times 10\times {{(0.5)}^{2}}$

$\displaystyle =0+(5\times 0.25)\,\,\,=\,1.25\,\mathbf{m}$

Question: An object is thrown vertically upwards and rises to a height of 10m. Calculate

(i) the velocity with which the object was thrown upwards and

(ii) the time taken by the object to reach the highest point.

Solutions: Distance traveled $\displaystyle (h)=10\,\mathbf{m}$

Final velocity $\displaystyle (v)=0\,\mathbf{m/s}$

Acceleration due to gravity, $\displaystyle g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

Acceleration of the object, $\displaystyle a=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

(i) $\displaystyle {{v}^{2}}={{u}^{2}}+2gh$

$\displaystyle 0={{u}^{2}}+2(-9.8)\times 10$

$\displaystyle 0={{u}^{2}}-196$

$\displaystyle {{u}^{2}}=196$

$\displaystyle u=\sqrt{196}$

$\displaystyle u=14\,\mathbf{m/s}$

(ii) $\displaystyle v=u+gt$

$\displaystyle 0=14-9.8\times t$

$\displaystyle t=1.43\,s$

Question: To estimate the height of a bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 seconds to touch the water surface in the river. Calculate the height of the bridge from the water level $\displaystyle (g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})$ .

Solutions: Now, initial velocity of stone, $\displaystyle u=0$

Time taken, $\displaystyle t=2s$

Acceleration due to gravity, $\displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And, height of the bridge, $\displaystyle h=?$ (To be calculated)

We know that for a freely falling body

Height, $\displaystyle h=ut+\frac{1}{2}g{{t}^{2}}$

Putting the above values in this formula, we get

$\displaystyle h=0\times 2+\frac{1}{2}\times (-9.8)\times {{(2)}^{2}}$

or $\displaystyle h=\frac{1}{2}\times 9.8\times 4$

or $\displaystyle h=-19.6\,\mathbf{m}$

Question: When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, $\displaystyle g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$ )

Solutions: Here, Initial velocity of ball, $\displaystyle u=?$ (To be calculated)

Final velocity of ball, $\displaystyle v=0$

Acceleration due to gravity, $\displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And height, $\displaystyle h=19.6\,\mathbf{m}$

Now, putting all these values in the formula:

$\displaystyle {{v}^{2}}={{u}^{2}}+2gh$

We get $\displaystyle {{(0)}^{2}}={{u}^{2}}+2\times (-9.8)\times 19.6$

$\displaystyle 0={{u}^{2}}-19.6\times 19.6$

$\displaystyle {{u}^{2}}={{(19.6)}^{2}}$

So, $\displaystyle u=19.6\,\mathbf{m/s}$

Here, Final velocity, $\displaystyle v=0$ (The ball stops)

Initial velocity, $\displaystyle u=19.6\,\mathbf{m/s}$ (Calculated above)

Acceleration due to gravity, $\displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And, Time $\displaystyle t=?$ (To be calculated)

So, putting these values in the above equation, we get

$\displaystyle 0=19.6+(-9.8)\times t$

$\displaystyle 0=19.6-9.8t$

$\displaystyle 9.8t=19.6$

$\displaystyle t=\frac{19.6}{9.8}$ $\displaystyle t=2s$

Thus, the ball takes 2 seconds to reach the highest point of its upward journey. Please note that the ball will take an equal time that is 2 seconds to fall back to the ground. In other words, the ball will take a total of 2 + 2 = 4 seconds to reach back to the thrower.

Question: A cricket ball is dropped from a height of 20 metres.

(a) Calculate the speed of the ball when it hits the ground.

(b) Calculate the time it takes to fall through this height. $\displaystyle (g=10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})$

Solution: (a) Here, initial speed, $\displaystyle u=0$

Final speed, $\displaystyle v=?$ (To be calculated)

Acceleration due to gravity, $\displaystyle g=-10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And, Height, $\displaystyle h=10\,\mathbf{m}$

Now, we know that for a freely falling body,

$\displaystyle {{v}^{2}}={{u}^{2}}+2gh$

So, $\displaystyle {{v}^{2}}={{(0)}^{2}}+2\times (-10)\times 20$

$\displaystyle {{v}^{2}}=-400$

$\displaystyle v=-\sqrt{400}$

or, $\displaystyle v=-20\,\mathbf{m/s}$

Thus, the speed of cricket ball when it hits the ground will be 20 metres per second. The minus sign with speed (or velocity) shows that it is in the downward direction.

(b) Now, initial speed, $\displaystyle u=0$

Final speed. $\displaystyle v=-20\,\mathbf{m/s}$ (Calculated above)

Acceleration due to gravity, $\displaystyle g=-10\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And, Time, $\displaystyle t=?$ (To be calculated)

Putting these values in the formula:

$\displaystyle v=u+gt$ ,

We get: $\displaystyle -20=0+(-10)\times t$

$\displaystyle -20=-10t$

$\displaystyle 10t=20$

$\displaystyle t=\frac{20}{10}$

$\displaystyle t=2s$

Thus, the ball takes 2 seconds to fall through a height of 20 metres.

Question: A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? $\displaystyle (g=9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}})$ .

Solution: Here, initial speed of ball, $\displaystyle u=15\,\mathbf{m/s}$

Final speed of ball, $\displaystyle v=0$ (The ball stops)

Acceleration due to gravity, $\displaystyle g=-9.8\,\mathbf{m/}{{\mathbf{s}}^{\mathbf{2}}}$

And, Height, $\displaystyle h=?$ (To be calculated)

Now, putting all these values in the formula,

$\displaystyle {{v}^{2}}={{u}^{2}}+2gh$

We get, $\displaystyle {{(0)}^{2}}={{(15)}^{2}}+2\times (-9.8)\times h$

$\displaystyle 0=225-19.6\,h$

$\displaystyle 19.6h=225$

or $\displaystyle h=\frac{225}{19.6}$

$\displaystyle h=11.4\,\mathbf{m}$

Thus, the ball will go to a maximum height of 11.4 metres before it begins to fall.

CBSE Tutorials

Gravitation | Equation of Motion of freely falling Bodies | Class 9

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Gravitation | Equation of Motion of freely falling Bodies | Class 9

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