A stone is thrown out from the window of a building. The stone is given an initi

Question

A stone is thrown out from the window of a building. The stone is given an initial speed of 8.00 m/s at an angle of 20.0 degree below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the stone hit the ground? (b) Find the height from which the stone was thrown. (c) What are the horizontal and vertical components of the velocity with which the stone hits the ground? (d)What is the final speed of the stone just before it hits the ground?

Explanation / Answer

Here ,

u = 8 m/s

theta = 20 degree

t = 3 s

a) horizontal distance = u * cos(theta) * t

horizontal distance = 8 * cos(20 degree) * 3

horizontal distance = 23.8 m

b)

let the initial height is h

h = 8 * sin(20 degree) * 3^2 + 0.50 * 9.8 * 3^2

h = 68.7 m

the height of building is 68.7 m

c)

at the ground

horizontal velocity = 8 * cos(20 degree) = 7.51 m/s

vertical velocity = - 8 * sin(20) - 9.8 * 3 = - 32.2 m/s

d)

final speed of the stone = sqrt(7.51^2 + 32.2^2)

final speed of the stone = 33 m/s

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