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A ball is thrown from \(1 \mathrm{~m}\) above the ground. It is given an initial velocity of \(20 \mathrm{~m} / \mathrm{s}\) At an angle of 40 degrees above the horizontal Find the maximum height reached and velocity at that point
in Physics & Chemistry by Platinum (122,256 points)
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x(t)=v \cos (\theta) t=20 \cos (40) t=15.3 t \\
y(t)=y_{0}+v \sin (\theta) t-\frac{9.8 t^{2}}{2} \\
y(t)=1+20 \sin (40) t-4.9 t^{2} \\
y(t)=1+12.9 t-4.9 t^{2} \\
v_{y}(t)=v \sin (\theta)-9.8 t \\
v_{y}(t)=12.9-9.8 t \\
\max \text { heightat } v_{y}(t)=0 \\
12.9-9.8 t=0 \\
-9.8 t=-12.9 \\
t=\frac{-12.9}{-9.8}=1.3 \\
y(1.3)=1+12.9(1.3)-4.9(1.3)^{2} \\
y(1.3)=9.5 m
\(y\) component of velocity is 0 at highest pt total velocity \(=v_{x}=15.3 \frac{\mathrm{m}}{\mathrm{s}}\)
by Platinum (122,256 points)

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