A 0.27 kg rock is projected from the edge of the top of a building with an initial velocity of 11.7 m/s at an angle 58◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.9 m from the base of the building.

Question

German Sharp

Physics

28 January, 06:32

A 0.27 kg rock is projected from the edge of the top of a building with an initial velocity of 11.7 m/s at an angle 58◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.9 m from the base of the building.

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Zaire Heath · Answer 1

Incomplete question. The complete question is

A 0.27 kg rock is projected from the edge of the top of a building with an initial velocity of 11.7 m/s at an angle 58 above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 15.9 m from the base of the building. How tall is the building? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s

Answer:

Total height of the building = 6.78 m

Explanation:

v = 11.7 m/s

Horizontal component of velocity = v x cos 58 = 11.7*cos 58 = 6.2 m/s

Total time of flight = Horizontal distance / Horizontal component of velocity = 15.9 / 6.2 = 2.56 sec.

Vertical component of velocity = v x sin 58 = 11.7 x sin 58 = 9.92 m/s

Time taken to reach maximum height = Vertical component of velocity / g = 9.92 / 9.81 = 1.01 sec.

Hence, Time taken to reach ground from maximum height = 2.56 - 1.01 = 1.55 sec.

Hence, Maximum height from ground = 0.5 x g x t^2 = 0.5 x 9.81 x 1.55^2 = 11.78 m

Maximum height from the top of building = 0.5 x 9.81 x 1.01^2 = 5 m

Hence, Total height of the building = 11.78 m - 5 m = 6.78 m