Question

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!

Answer 1

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: