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March 31, 2015

March 31, 2015

Posted by **John** on Friday, May 30, 2014 at 10:22pm.

- Physics -
**MathMate**, Saturday, May 31, 2014 at 7:02am(a)

KE=(1/2)mv²

(b)

At maximum height,

KE in y-direction = 0, i.e.

loss in KE

=(1/2)m(v sin(θ))²

(c)

Maximum height, h

equate mgh to loss in KE, or

mgh = (1/2)m(v sin(θ))²

Solve for h.

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