Home 1st Year Physics Notes FSC 1st Year Physics Chapter 3 Exercise & Numericals Notes

# FSC 1st Year Physics Chapter 3 Exercise & Numericals Notes

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In this post, I am sharing FSC 1st Year Physics Chapter 3 solved Exercise & Numericals Notes PDF for the students of 11th Class. This is the 3rd chapter in FSC 1st Year Physics book and its name is Motion and Force. Students can download these 11th class physics Chapter 3 Motion and Force Solved Exercise and Numbericals notes PDF on their laptop or mobile. These Physics Notes for class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. On ilmihub.com, you can get the complete Class 11 Physics Notes PDF of all chapters for free.

## FSC Part 1 Physics Chapter 3 Motion and Force Solved Exercise Numericals Notes PDF Download

### A helicopter is ascending vertically at the rate of 19.6ms-1. When it is at the height of 156.8m above the ground, a stone dropped. How does the stone take to reach the ground?

solution;
data;
speed of helicopter = 19.6 ms-1
height of helicopter = 156.8m
when a stone dropped, it moved upward due to inertia
initial velocity of stone = 19.6 ms-1
Final velocity = vf = 0 , a = g = 9.8ms-2
To Find
time = t = ?
for upward distance = t1 = ?
Calculation
time to reach from A to C
vf = vi + gt1

0 = 19.6 – (9.8 x t1)
0 = 19.6 – 9.8 x t1
9.8 x t1 = 19.6
t1 = 19.6/9.8
t1 = 2s
distance moved in the upward direction in time t1 is S1 = ?
S1 = vt1 + ½ gt1 2
= 19.6 x2 + ½ (-9.8)(2)2
= 39.2 – 19.6
S1 = 19.6m
total height from the ground = S = CD+BD = 156.8+19.6 = 176.4m
Time from C to B = t2
vi = 0, a =g = 9.8
S = 176.4 m
S2 = v21 + ½ gt2 2
176.4 = 0+1/2 x 9.8 x t22
176 = 4.9t22
t22  = 176.4/4.9 =36
t2 = 6sec
required total time = t1 + t2
t       = 2+6
t = 8 sec

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