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A slingshot fires a pebble from the top of a building at a speed of 14.0 m/s. The building is 33.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

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Answer:

Its vertical V = sqrt.(2gh) = sqrt.(19.6 x 31) = 24.65m/sec.

a) Sqrt.(24.6^2 + 13^2) = 27.8m/sec.

Time to max. height = (v/g) = (13/9.8) = 1.326 secs.

Height gain = 1/2 (t^2 x g) = 8.6 metres.

b) V at ground = sqrt.(2gh) = sqrt.(19.6(31 + 8.6)) = 27.86m/sec.

c) Since it will be going 13m/sec. in (b) as it falls to the building height, the V at the ground will be the same (27.86m/sec.).

Explanation:

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