Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Please show work!

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:

real, smaller and upside-down.   so the answer should be b.