What is the answer for question 3.b)?

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Let board making angle `theta` with the horizontal .In t=2 seconds rolled over a distance S = 1.3 m.

We ignore the size of the ball, which can slow down the ball.

Let g=acceleration due to gravity = 9.8 m/s²

then

a=acceleration down the board

=g sin(θ)

u=initial velocity=0

S=1.3 m

t = 2 sec

then

`s=ut+(1/2) g t^2 sin(theta)` (i)

1.3=0+2x9.8xsin(`theta`)

sin(`3.8^o`

differentiate (i) with respect to t

`(ds)/(dt)=u+g t sin(theta)`

But `(ds)/(dt)=speed =v` (say)

put u=0

`v=g t(.0663)` , put value of g

v=.65t

v alog y -axis and t along x-axis.

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