283height of projectile B at 1/2 second= 10 *1/2 - 1/2*10*1/4=3.75 m
For projectile A=
H=20sina*1/2 - 1.25
3.75+1.25=10sina
1/2=sin a a=30°
Got the second method :-D
3 Answers
206Resolve vA into its horizontal and vertical components and then equate the height at which they meet.......from here you get angle, then use horizontal component to find x
- Sahil Jain i thought we can do it with relative velocity of 2 projectileUpvote·0· Reply ·2013-06-08 08:29:46
- Sahil Jain Sir can u please explain ur method.
283Uay=20sina and Uby=10
Uax=20cosa and Ubx=0
aax= -g
abx= -g
Relative acceleration= 0
SIne the relative acceleration is zero the relative motion will be uniform. So
Uay=Uby
20sina=10
sina=1/2
a=30°
R=Uax*t=20cos30 * 1/2=5√3
283