Question

Three particles A,B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally .They hit the ground with speeds $v_A,v_B$ and $v_C$ respectively. Then,

• A. $v_A=v_B=v_C$
• B. $v_A=v_B>v_C$
• C. $v_A>v_B>v_C$
• D. $v_A>v_B=v_C$

Answer 1

The correct option is A $v_A=v_B=v_C$

For A: it goes with velocity u will reach its maximum height(i.e velocity becomes zero) and comes back to O and attain velocity u.
Using $v^2=u^2+2as\Rightarrow v_A=\sqrt{u^2+2gh}$
For B: going down with velocity u:
$v_B=\sqrt{u^2+2gh}$
For C: horizontal velocity remains same i.e u.Vertical velocity $v_y=\sqrt{0+2gh}=\sqrt{2gh}$
The resultant, $v_c=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+2gh}$
Hence, $v_A=v_B=v_C$