Question

-) A ball is projected at an angle of 45° to the horizontal. If the horizontal range is 20 m, the maximum height to which the ball rises is (g 10 ms 3). (a) 2.5 m (bi 5.0 m (​

Answer 1

Answer:

Explanation:

Range = (Vi^2 * sin2θ) / g where g = 9.8 m/s^2 and θ = 45 degrees

Maximum Height = (Vi^2 * sin^2 θ) / 2g

Using the range formula

Range = (Vi^2 * sin2θ) / g

20 m = (Vi^2 * sin 2 * 45) / g

20 m = (Vi^2 ) * sin 90) / g

20 m = Vi^2 * 1) / g

20 m = Vi^2 / g

Solving for Vi^2

Vi^2 = 20 m * g

Solving for the maximum height, maxH of the ball

maxH = (Vi^2 * sin^2 45) / 2g

maxH = (Vi^2 * 0.5) / 2g

maxH = (20 m * g * 0.5) / 2 g

maxH = (10 m * g) / 2g

maxH = 5 m

The maximum height reached by the ball is 5 meter

Answer 2

Answer:

it is maxH=5.0m.

Explanation:

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