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What is the answer for question 3.b)? http://postimg.org/image/q8ectggh3/

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lkballer24 | Student, Grade 11 | Valedictorian

Posted April 19, 2013 at 2:32 PM via web

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What is the answer for question 3.b)?

http://postimg.org/image/q8ectggh3/

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pramodpandey | College Teacher | Valedictorian

Posted April 20, 2013 at 12:38 PM (Answer #1)

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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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