A ball is acted upon by the following velocities. (i) 3 ms^(-1) due East 9ii) 11 ms^(-1) due South and (iii) 5 sqrt 2 ms^(-1) due North-East. Find the magnitude and direction recultant velocity.
Answers
Explanation:
Solution :
The v arious velocities acting on the ball are shown in Fig. 2 (c ) .59 (a0. →OP=3ms-1
due East, →OQ=1ms-10 due South an vec (OS) =5 sqrt 2 ms^(-1)∈cl∈edatan∠45^(0)withE∗.Resolv∈gvec (OS)` into two rectangular components along East and North directions, we get
.
5√2cos450=5ms-1 along East
and 5√2sin450=5ms-1 along Northe
:. Effective velocity of ball along East,
A=3+5=8m/s, represented by →OP′
and the effective velcoity of ball along South,
B=11-5=6m/s, represented by→OQ′.
The resultant velcity →R will be represented by →OT, where ∠P′OQ′=900. Fig. 2 (c ) .59 (b).
Thus magnitude of →R will be R= (sqrt A^(2) +B^(2) ) =sqrt(8^(2) +6^(2)) + =10 ms^(-1)Letvec Rmakeand∠betawithE∗,then
tanbeta =(B)/(A) =6/8 =0.75 =tan =tan 36 &(0) 52'orbeta =36^(0) 52' South of East `.
Answer:
We break the velocity V3 into its component
Net velocity along south, VS = V2 – V3N
= 11 – 5 = 6 m/s
Net velocity along east, VE = V1 + V3N
= 3 + 5
= 8 m/s
So,
So, magnitude of resultant velocity
Let, θ be the angle which the resultant makes with east direction