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Worked Examples-Calculation Part 2 - Motion in a Straight Line

Considered Formula; Motion Under Gravity

i)v=u+gt
Where  uInitial velocity,vFinal velocityaAcceleration due to gravitytTime
ii)s=ut+12gt2
Where  sDistance traveled,uInitial velocity,aAcceleration due to gravitytTime
iii)v2=u2+2gs
Where  uInitial velocity,vFinal velocity,aAcceleration due to gravity,sDistance traveled

Example 1

A stone of mass 2kg is released from a height of 20m above the ground. Find

a)    Velocity reached just before hit the ground

b)    Time taken to reach the ground

Solution

Data given

Distance (s) = 20 m

Accelaration due to gravity (g) = 10 m/s2

a)        Final velocity (v) = ?

                                      v2=2gsInitial velocity (u)=0

                                       v=2gs
                                         =2(10)(20)
                                         =400
                                         =20 m/s

          The final velocity = 20 m/s

b)        Time (t) = ?

                                        v=gt

                                      20=10t
                                      10t10=2010
                                         t=2 sec

          The time taken to reach the ground = 2seconds

Example 2

A stone is thrown vertically upward from the ground with a velocity of 30 m/s. find

a)    Maximum height reached

b)    Time taken to reach the maximum height

c)    Time taken to reach the ground again

d)    The velocity reached half-way to the maximum height

Solution

Data given

Initial velocity (u) = 30 m/s

Final velocity (v) = 0 m/s

Accelaration due to gravity (g) = 10 m/s2

a)        Distance (s) = ?
    For upward motion, g = -10 m/s2

                                      v2=u2+2gs

                                      02=302+2(10)s
                                        0=90020s
                                    20s=900
                                   20s20=90020
                                         s=45 m

          The maximum height reached = 45 m

b)        Time taken to reach maximum height (t1) = ?

                                        v=u+at

                                      0=30+10t1
                                       10t1=030
                                       10t1=30
                                      10t110=3010
                                         t1=3 sec

          The time taken to reach maximum height = 3 seconds

c)        Time taken to reach ground again (t2) = ?

                                        t2=2t1

                                            =2×3
                                            =6 sec

          The time taken to reach ground again = 6 seconds

d)        Final velocity (v) = ? Time (t)=(32)=1.5 sec

                                        v=u+at

                                            =30+(10×1.5)
                                            =3015
                                            =15 m/s2

          The velocity reached half-way to the maximum height = 15 m/s

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