A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?

Initial speed u = 29.94 m/s or 98.23 ft/s

Maximum height h = 22.87m or 75.03ft

Step-by-step explanation:

The range of a projectile is the horizontal distance covered by the projectile. And can be written as;

Range R = u^2sin2θ/g

Given:

R = 300 ft = 91.44 m

Angle θ = 45°

u = initial speed

g = acceleration due to gravity = 9.8m/s^2

u^2 = Rg/sin2θ

u = √(Rg/sin2θ)

u = √(91.44×9.8/sin(2×45))

u = 29.94 m/s or 98.23 ft/s

The maximum height can be derived from

h = u²sin²θ/2g

Substituting the values.

h = (29.94)^2 × sin²(45) / (2×9.8)

h = 22.87m or 75.03ft

x/3-10=8

x/3=18

x=48

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