Askateboarder, with an initial speed of 2.0 m/s, rolls virtually friction free down a straight incline of length 18 m in 3.3 s. at what angle θ is the incline oriented above the horizontal?

Initial speed of the skateboarder (u) = 2 m/s

Distance covered (s) = 18 m

Time taken = 3.3 seconds

Let the acceleration be a.

Using seconds equation of motion:

a = 2.09 m/s^2

Now, Acceleration down the incline = g Sin Θ

g Sin Θ = a

9.8 × Sin Θ = 2.09

Sin Θ =

Θ = 12.31°

Hence, the angle of the inclined plane is: Θ = 12.31°

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

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