Question

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Answer 1

7.We have

intial velocity=u m/s
final velocity=0 m/s

Acceleration due to gravity= - 10m/s^2

According to third equ. if motion

2as=v^2-u^2
2as=0^2-u^2
-20s= -u^2
s=-u^2/-20
s=0.05u^2

8.We have,

initial velocity=u m/s
acceleration= a m/s^2

i.Time=1sec

S=ut+1/2at^2
S=u+1/2a
S=u+0.5a

ii.u=2m/s
a=1 m/s^2
t=5sec

S=ut+1/2at^2
S=2(5)+0.5(1)(5)^2
S=10+12.5
S=22.5m

9. Distance travelled=Area of figure OABC
=Area of rectangle OADC + Area of triangle ABD

We will now find out the area of the rectangle OADC and the area of the triangle ABD.
(i) Area of rectangle OADC=OA × OC=u × t=ut ...... (5)
(ii) Area of triangle ABD=(1/2) × Area of rectangle AEBD
=(1/2) × AD × BD
=(1/2) × t × at (because AD 
= t and BD = at)=(1/2) at2...... (6)

So, Distance travelled, s=Area of rectangle OADC + Area of triangle ABDor s = ut + (1/2) at2
This is the second equation of motion. It has been derived here by the graphical method.

10.Linear motion /uniform motion