Question

a shell is projected with speed 150m/s at 37 with horiZontal.. at highest point it shoots 1/4 of mass with 250 m/s speed. the horizontal range of remaining part is 1770 how?​

Answer 1

Answer :- V' = 3V cos theta

Explanation:

mV cos theta =-mV cos theta /2 + m/2 ×V'

i.e. (1+1/2) mV cos theta = m/2 V'

i.e. V' =3V cos theta