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qwelly [4]
3 years ago
12

A soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill

Physics
1 answer:
ozzi3 years ago
4 0

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

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Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

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Given;

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h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

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Answer:

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