Posted by nady on .
2. A bullet is fired from the ground at an angle of 45o above the horizontal. What initial speed vo must the bullet have in order to hit a point 550 ft high on a tower located 600 ft away (ignoring air resistance)?
Y^2 = Yo^2 + 2g*d = 0.
Yo^2 - 64*550 = 0.
Yo^2 = 64*550 = 35,200.
Yo = 187.6 Ft/s = Ver. component of Vo.
Vo = Yo/sinA=187.6 / sin45=265.3 Ft/s. = Initial velocity.