If an object is launched straight up into the air from a starting height of h_{0} feet, then the height of the object after t seconds is approximately h=-16t^2+v_{0}t+h_{0} feet, where v_{0} is the initial velocity of the object. Find the starting height and initial velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

Answer 1

V = Vo + gt = 0.
Vo - 32*5 = 0.
Vo = 160 Ft/s.

h = ho + Vo*t + 0.5g*t^2 = 412 Ft.
ho + 160*5 - 16*5^2 = 412.
ho + 800 - 400 = 412.
ho + 400 = 412.
ho = 412 - 400 = 12 Ft.

Answer 2

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)?