from the base of lamp post, a particle is projected with velocity 10m/s at the angle 60 degree with horizotal. the height of lamp post is 4 times the maximum hight attained by particle. when the particle is at highest point the motion of it's shadow is observed on x-axis. then find acceleration of shadow, velocity of shadow , velocity of shadow with respect to particle and time at which particle and shadow coincide
challenge for goiit'ans
See the acceleration of the shadow will be Zero as we know the shadow moves in forwqrd direction (along x axis)and a projectile 's motion in horizontal direction is not accelerated and the velocity in X axis thus remains constant
Correct me if i am wrong
dis was d solution dat i posted when u asked dis question earlier...
using equations of projectile motion for range, height and time of flight...
we get... height of lamp = 15m... distance travelled by d shadow by d tym particle reaches d max. ht. position = 10/sqrt(3) ... and d time of ascent = (sqrt3) / 2...
now using d equations of motion... we get
v = 20/3 + [(sqrt3)/4].a and u = 20/3 - [sqrt(3)/4].a ...
vel of shadow = 20/3 m/s... vel. of shadow wrt particle = 20/3 - 10(cos60) = 5/3 m/s ... and both d shadow n d particle coincide when d particle touches d ground... hence d tym taken = time of flight = sqrt(3) sec...
i think d acceleration of d shadow will be zero since it depends only on d horizantal motion of d particle and d particle has constant vel along horizantal direction... hence acc must be zero...!!!!! yet... i'm not sure of dat... but d given 3 ans are correct...
rate if useful...
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