A bullet is fired from the ground at an angle of 60^o above the horizontal. What initial speed 𝑣𝑣0 must the bullet have in order to hit a point 500 ft high on a tower located 800 ft away (ignoring air resistance) ?

Answer 1

Heavens, we are using feet ?
I guess the g = 32 ft/s^2 approximately if I remember from before the attack on Pearl Harbor.
Divide this into a vertical problem and a horizontal problem
I am going to call initial speed = s
Vertical
initial speed up = Vi = s sin 60 = .866 s
v = .866 s - 32 t
h = 0 + Vi t - 16 t^2
500 = .866 s t - 16 t^2
16 t^2 - .866 s t + 500 = 0
Horizontal
u = S cos 60 forever = 0.5 s
d = u t = 0.5 s t = 800
s t = 1600
t = 1600/s
---------------------------------now plug and chug
16 (1600^2/s^2) - .866 s(1600/s) + 500 = 0
16 (1600^2/s^2) - 1386 + 500 = 0
16 (1600^2/s^2) = 886
s^2 = 46230
s = 215 ft/second
Always check my arithmetic carefully :)