IIT JEE , AIEEE Forum - Physics , Mechanics

Mr.IITIAN007

A particle is projected from a point P(2,0,0) with a velocity 10m/s making an angle 45 degrees with the horizontal . The plane of projectile motion
through a horizontal line PQ which makes an angle of 37 degrees with the +ve X-axis, XY-plane is horizontal .The coordinates of the point where
the particle will strike the line PQ is (taking g=10m/(sec)2):-
(a) (10,6,0)
(b) (8,6,0)
(c) (10,8,0)
(d) (6,10,0)

pinakinab

is it b) ?

deepti_hari

i think the answer is a) 10,6,0.the diagram is given in the form of attatchment.

explaination is as follows

given that P(2,0,0) and u = 10m/sec

and the angle made by P Q with =ve X axis is 37 degrees

now range of the projectile is given by u*u sin(2*45)/g

= 10*10*1/10 = 100/10 = 10units hence PQ = 10units

now using parametric form of line formula

Q(2+10cos 37,0+10sin 37,0) because it is in XY plane z co-ordinate will always be zero.

By simplifying we get q(10,6,0)

sanchay_1991

ans is (a) only

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