# Physics

### Chapter : 1. Motion

#### Examples on Body Falling Freely Under Gravity

Example : 13
A train is moving with a velocity 400 m/s. With the application of brakes a retardation of 10 m/s2 is produced. Calculate the following :
(i) After how much time it will stop ?
(ii) How much distance will it travel before it stops?
Solution. (i) Given: u = 400 m/s, a = –10 m/s2, v = 0, t = ?
Using equation, v = u + at, we get
0 = 400 + (–10) × t
or t = 40 s
(ii) For calculating the distance travelled, we use equation,
v2 = u2 + 2as, we get
(0)2 = (400)2 + 2 × (–10) × s
or 20s = 400 × 400
or s = 8000 m = 8 km

Example : 14
A body is thrown vertically upwards with an initial velocity of 19.6 m/s. If g = –9.8 m/s2. Calculate the following :
(i) The maximum height attained by the body.
(ii) After how much time will it come back to the ground ?
Solution. (i) Given: u = 19.6 m/s, g = –9.8 m/s2, v = 0, h = ?
Using equation v2 = u2 + 2gh, we get
(0)2 = (19.6)2 + 2(–9.8) × h
or h = = 19.6 m
(ii) Time taken to reach the maximum height can be calculated by the equation,
v = u + gt
or 0 = 19.6 + (–9.8) × t
ort = 2s
In the same time, it will come back to its original position.
Total time = 2 × 2 = 4s

Example : 15
From the top of a tower of height 490 m, a shell is fired horizontally with a velocity 100 m/s. At what distance from the bottom of the tower, the shell will hit the ground ?
Solution. We know that the horizontal motion and the vertical motion are independent of each other. Now for vertical motion, we have u = 0, h = 490 m, g = 9.8 m/s2, t = ?
Using equation,h = ut +gt2, we get
490 = 0 + × 9.8 × t2
or t2 = 490/4.9 = 100
or t = 10 s
∴ It takes 10 seconds to reach the ground.
Now, horizontal distance = horizontal velocity × time
= 100 m/s × 10 s = 1000 m
∴ The shell will strike the ground at a distance of 100 m from the bottom of the tower.