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The roof of a two-story house makes an angle of 30°with the horizontal. A ball rollingdown the roof rolls off the edge at a speed of 5.0m/s. The distance to the ground fromthat point is 7.0 m. (a) How long is theball in the air? (b) How far from thebase of the house does it land? (c) What is its speedand direction just before landing?

User Noremac
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1 Answer

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Answer

angle made by the horizontal = 30°

velocity = 5 m/s

distance = 7 m

Vertical component (V y) = 5 sin (30°) = 2.5 m/s

Horizontal component(V x) = 5 cos (30° ) = 4.33 m/s

Applying


s = u t +(1)/(2)gt^2


7 = 2.5 t +0.5*9.8 t^2

t = 0.967 s

b) Distance from base

=
V_x * t

=
2.5* 0.967

= 2.42 m

c) You can find final vertical velocity

V= u + gt

V = 9.8 × 0.967

v = 9.48 m/s

Final velocity =
√((9.48^2 - 4.33^2))

v = 8.433 m/s

Final angle


\theta = tan{-1}(v_y)/(v_x)


\theta = tan{-1}(9.48)/(4.33)

θ = 65.41°

User Sai Anudeep
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