TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, bodies are thrown with speed u1 and u2,
i.e. u1=u2=10ms−1.
One is vertically up, i.e. 90∘ with horizontal, and other is thrown at angle θ with vertical,
i.e. θ=60∘.
If required time, t=2s.Now, s12 be the relative position of 1 with respect to 2 and v12 be the relative speed of 1 with respect to 2 .
As we know that,
Formula of resultant vector v1 and v2 by using triangle law,
v12=√v12+v22−2v1v2‌cos‌θ
=√102+102−2(10)2×cos‌60∘
=√200−200×‌

1

2

=√100
=10
and ‌‌v12=‌

s12

t

∴‌‌s12=v12t
=10×2
=20m