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A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45 degrees. With what speed was it thrown?

User ManneR
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1 Answer

4 votes

Answer:

Step-by-step explanation:

This is the case of horizontal projection from a height:

Time, t = sqrt ( 2h / g )

= sqrt ( 2 * 20 / 9.8 )

= 2.02 s

Vfx = V

Vfy = g* t = 2.02 g

theta (θ)= 45 deg

tan theta (tan θ) = Vfy / Vfx

tan 45 = 2.02 g / V

V = 2.02 * 9.8

= 19.8 m/s

≅ 20m/s

User Jan Koch
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