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# A particle is projected from a point(0,1) on Y-axis aiming towards a point(4,9). It fellson ground along X-axis in 1 second. Taking g=10m/second square , and all coordinate in metres, find the X- coordinatewhere it fell

Umesh Jakhar
32 Points
10 years ago

tanΦ=9-1/4-0=2=perpendicular/base.      ...........(1)

where Φ is angle of projection

given that time of flight is 1 second

from s=ut-1/2gt2

(-1)=usinΦ(1)-1/2(10)(1)   final displacement is -1 when it hit the ground ok!!

usinΦ=4

then

ucosΦ=usinΦ/tanΦ=4/2=2    (from equestion (1) tanΦ=2 )

then x coordinate = VxT=(2)X(1)=2   (Vx = horizontal componant of velocity)

particle hit the ground at (2,0) OK!!!!!!!!!

GOOD LUCK

you will crack JEE

DON't forgate to approve my answer this will be by clicking yes below... ok!!!!

Vibekjyoti Sahoo
145 Points
6 months ago

tanΦ=9-1/4-0=2=perpendicular/base.      ...........(1)

where Φ is angle of projection

given that time of flight is 1 second

from s=ut-1/2gt2

(-1)=usinΦ(1)-1/2(10)(1)2     final displacement is -1 when it hit the ground ok!!

usinΦ=4

then

ucosΦ=usinΦ/tanΦ=4/2=2    (from equestion (1) tanΦ=2 )

then x coordinate = VxT=(2)X(1)=2   (Vx = horizontal componant of velocity)

particle hit the ground at (2,0)