The ratio of the ball's maximum height to its range is 0.25
Let u represent the initial velocity of the projectile, g is the acceleration due to gravity and θ is the angle.
The maximum height (h) is:
[tex]h=\frac{u^2sin^2(\theta)}{2g} \\\\h=\frac{u^2sin^2(45)}{2g}=0.25(\frac{u^2 }{g})[/tex]
The range (R) is:
[tex]R=\frac{u^2sin(2\theta)}{g} \\\\R=\frac{u^2sin(2*45)}{g}=(\frac{u^2 }{g})[/tex]
The ratio of the ball's maximum height to its range = [0.25u²/g]/[u²/g] = 0.25
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