A projectile is projected with a velocity of $20 m s^{- 1}$ making an angle of $60^{0}$ with the horizontal. Find its average velocity between the instant it crosses half the maximum height (in m/s)

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Detailed Solution

$V_{x} = \frac{d x}{d t} = 3$ and
$V_{y} = \frac{d y}{d t} = 4 - 10 t \Rightarrow V_{y} = 4 - 10 (0) = 4 V = {[V_{x}^{2} + V_{y}^{2}]}^{1 / 2} V = {[3_{}^{2} + 4_{}^{2}]}^{1 / 2} ∴ V = 5 m / s$

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