114 - Kinematics Questions Answers
a cyclist is moving due east with a velocity of 10 km/h there is no wind air and rain appear to fall at degree vertival calculate the speed of rain.
Question is not submitted properly, angle is not given
let the angle is θ degree
then v(man) = 10i
and v(rain) = -vj i and -j are the directions for man and rain
so v(rain w.r.t. man) = -vj-10i
so tanθ = 10/v so v = 10 cotθ
θ is measured from vertical
A body is projected with a velocity 'u' making an angle α with the horizontal. Its velocity when it is perpendicular to the initial velocity vector is:
A. ucotα B. utanα C. usinα D. ucosα
component of gravity perpendicular to the final direction is gsinθ so by equation
v=u+at
0 = u -gsinθt
so t = u/gsinθ
now along final direction v = 0+gcosθt on putting the value of t we get v = gcosθu/gsinθ = ucotθ
Due to gravity a freely falling body covers the distances in the ratio 1:3:5:7:9..............................so
if a particle passes at the same point twice (2nd and 4th sec), then it will be at heightest pont in 3rd sec. so height = 5+15+25 = 45 meter
1>What is the distance covered by a point object projected with velocity "U" making an angle theta with the horizental?
formula for calculating distance of a curve is ∫ds
where ds2 = dx2 + dy2
and equation of projectile is
y = xtanθ-(1/2)(gx2/u2cos2θ)
try to solve
if not solved give qurey i will give the complete solution
A motorboat going downstream, passes a raft at point 'A'.t = 60 min later it turned back and after some time it passes the same raft at a distance of 6 k.m. from point 'A' . find the flow velocity assuming the duty of the engine to be constant.
you can see video in you tube for this irodov problem 1.1
a body is projected at 40m/s from the horizontal ground at an angle 60 degree with the vertical. time at which it is moving perpendicular to its direction of motion is
when particle will move perpendicular to its initial direction its velocity in its ititial direction will become 0. and component of gravity in that direction = gsin30
30 degree will be the angle of motion of particle from horizontal.
now use v = u + at, 
a particle is projected at an angle of 60 degree with the horizontal at a speed of 30m/s. the time after which the speed of the particle remain half of the initial speed is
intial speed = 30 m/s
at top point speed = 30 cos60 = 15
so answer will bo half time of flight
intial speed = 30 m/s
at top point horizontal speed = 30 cos60 = 15
vertical component will be 0
so speed = 15 m/s
so answer will bo half time of flight
= usinθ/g
a small block slides down a smooth inclined plane , starting from rest at time t = 0 .let Sn be the distance travelled by the block in time interval t = n - 1 to t = n. then the ratio Sn / Sn + 1 is ??
distance travelled in n seconds = 1/2 an2 (a is the acceleration in the direction of incline)
distance travelled in n-1 second = 1/2 a(n-1)2
distance travelled in n+1 second = 1/2 a(n+1)2
so according to given condition Sn / Sn+1 = [1/2 an2 - 1/2 an2] / [1/2 a(n+1)2 - 1/2 an2]
now solve
Projectile motions Plzz help!!!!?
2)If α = 30 degrees and β = 30 degrees and 'a'= 4.9 m, find the initial velocity of projection.
component of initial velocity along OB = ucos30
component of final velocity along OB = 0
component of g along OB = -gsin30
now use v = u + at
in second part clearly explain "a"
a body is projected vertically upward with speed 32 m/s. total time taken in its complete journey is
suppose it travelled h height thus from bottom to top it will covered h height in time t1 & at top its vel will be zero thus from eqn v= u-gt => 0= 32- 10t u can find t1= 3.2s , same time particle will take in downward journey so total time = 6.4 sec.
This answer is given by sarika and is correct answer, we made some modifications only