Answer:
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
The ball reaches the maximum height of 54 feet
The question is about projectile motion,
the ball is shot at an angle α = 45°, and
the initial velocity u = 46 ft/s.
Under the projectile motion, the maximum height H is given by:
[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]
where, g = 9.8 m/s²
substituting the given values we get:
[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]
Hence, the maximum height is 54 feet.
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