A ball is shot at an angle of 45 degrees into the air with initial velocity of 41 ft/sec. Assuming no air resistance, how high does it go?

The ball goes up to 43 ft high.

Explanation:

Initial velocity of the ball is 41 ft/sec

Angle of projection = 45⁰

Vi- up = 41 ft/sec X sin 45 = 41 X 0.707 = 28.987 ft/sec

Vi - horizontal = 41 ft/sec X cos 45 = 41 X 0.707 = 28.987 ft/sec

Time, t can be calculated as

t = 0m/s - 28.987 ft/s / - 9.8 ft/s²

t = 2.96 s

The average upward velocity is

Vup - ave = 28.987/2 = 14.5 ft/s

Therefore, height is

Height = ( Vup-ave ) (tup)

           = 14.5 ft/s X 2.96s

           = 43 ft

The ball goes upto 43 ft high.

the solution is 12x-6y