24P2Q01

1a)

$\displaystyle \begin{aligned}\text{gain in GPE}&=mg\Delta h \\27&=(0.43)(9.81)\Delta h \\\Delta h&=6.40\text{ m}\end{aligned}$

1b)

$\displaystyle \begin{aligned}({{v}^{2}}&={{u}^{2}}+2as) \\0&={{u}_{y}}^{2}+2(-9.81)(6.40) \\{{u}_{y}}&=11\text{ m }{{\text{s}}^{-1}} \end{aligned}$

1ci)

$\displaystyle \begin{aligned}{{u}_{y}}&=11 \\u\sin 32{}^\circ &=11\text{ } \\u&=20.76\text{ m }{{\text{s}}^{-1}} \end{aligned}$

$\displaystyle \begin{aligned}{{u}_{x}}&=u\cos 32{}^\circ \\&=20.76\cos 32{}^\circ \\&=17.60\text{ m }{{\text{s}}^{-1}} \end{aligned}$

By PCOM:
$\displaystyle \begin{aligned}(0.43)(17.60)+0&=(0.43+0.67)V \\V&=6.88\text{ m }{{\text{s}}^{-1}}\end{aligned}$

1cii)

The horizontal component of velocity as it rises is higher than when it falls.

Since the vertical height is the same, the time taken to rise and fall are the same.

Hence, the horizontal distance is higher as it rises.

xmPhysics