IIT JEE , AIEEE Forum - Physics , Mechanics

zeeshanp

two stones are projected from a point P with a velocity of 20m/s one at an angle @ above da horizon and the other at and angle @ to the verticle. find da distance of each stone from p one second after projection and da distance the stones are apart.(take cos@=4/5;sin@=3/5)

pls solve completely and ply cud u giv a diagram as well////

shub

for stone 1...
hor. velocity = v cos@
distance covered in 1 sec = v cos@ = 20*4/5 = 16m
vertical velocity = v sin@
distance covered in 1 sec = v sin@*1 - 1/2g1^2
= 20*3/5 - 5
= 7m
total distance = root(16^2 + 7^2)
= root(256 + 49)
root(305)

for stone 2.....
horiz velocity = v sin@ = 12
distance = 12
vertical velocity = v cos@
distance = vcos@*1 - 1/2gt^2
= 11
total distance = root(265)

for distance between the 2 stones....
we can consider the coordinates of A as (16,7)
and of B as (12,11)
distance between them = root(4^2 + 4^2)
=root 32

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