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Find the velo...

Question

Find the velocity of a projectile at the highest point, if it is projected with a speed $15 m s^{- 2}$ , in the direction $45^{\circ}$ above horizontal.
[take $g = 10 m s^{- 2}]$

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Solution

initially
$V_{y} (i n v e r t i c a l) = 15 s i n 45^{\circ} = \frac{15}{\sqrt{2}}$

$V_{K} (i n h o r i z o n t a l) = 15 c o s 45^{\circ} = \frac{15}{\sqrt{2}}$

$H e i g h t p o i n t V_{v} (i n h o r i z o n t a l = 15 c o s 45^{\circ} = \frac{15}{\sqrt{2}}$

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