23P2Q02

2a)

The speed is minimum at the top of the trajectory, when the speed is simply the horizontal component of velocity vx, because the vertical component of velocity is zero.

\displaystyle \begin{aligned}   {{{KE}_{min }}} &=65 \\  \frac{1}{2}(0.55){{v}_{x}}^{2}&=65 \\  {{v}_{x}} &=15.37\text{ m }{{\text{s}}^{-1}} \end{aligned}

\displaystyle v=\frac{{{v}_{x}}}{\cos 42{}^\circ }=\frac{15.37}{\cos 42{}^\circ }=20.69\text{ m }{{\text{s}}^{-1}}

\displaystyle {{v}_{y}}=20.69\sin 42{}^\circ =14\text{ m }{{\text{s}}^{-1}}

COMMENT: For a “show” question, start with an explanatory sentence, and show all substitution values explicitly.

2b)

Time to reach peak, \displaystyle {{t}_{p}}=\frac{{{u}_{y}}}{g}=\frac{14}{9.81}=1.427\text{ s}

Time of flight, \displaystyle {{t}_{f}}=2{{t}_{p}}

Horizontal range, \displaystyle R={{v}_{x}}{{t}_{f}}=(15.37)(2\times 1.427)=43.9\text{ m}

2c)

It takes shorter to reach its maximum height.

Air resistance acts in the same direction as gravity on the way up, but in opposite on the way down.

The object therefore decelerates at a faster rate on the way up than it accelerates on the way down.

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