A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high does it go

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

e/c, where e is its total energy