Answer:
Let’s assume that the projectile has the same initial velocity in both cases. It is necessary to find that velocity.
u = u[x]i +u[y]j = [u cos 15]i + [u sin 15]j
The p.v. ; s = s[x]i +s[y]j , get the time of flight by letting s[y]=0
s[y] = [u(y)].t -[1/2]gt^2 =0 , so t = 2u[y]/g
t = [ 2 u sin 15]/g eq 1
s[x] = [ u cos 15] .t eq 2 , substitute in eq q into eq2
range =r = [ u cos 15 ][ 2u sin 15 ]/g =[u^2/g][ 2 sin 15 . cos 15],
r =[ u^2 sin 30]/g =1.5 km
( from this we see that ,in general, r = [u^2 .sin (2a)]/g
u^2 = [ 3g] , eq 1
Now do the 2 nd part
r[2] = [u^2 sin (2a)]/g , eq 2 Let a =45 and substitute eq 1 into eq 2
r[2] = [3g sin 90]/g
r[2] =3 km.
Explanation:
