Question

Find maximum horizontal range when high is 25m and speed is 40ms-1

Answer 1

Height h = 25 m, u = 40 m/s, g = 9.8 m/s^2
$h \: = \: 2 5\: m \: = \: \frac{ { {u}^{2} \sin }^{2} \: \alpha }{2g}$
${ \sin}^{2} \: \alpha \: = \: \frac{2 \: \times \: 25 \: m \: \times \: 9.8 \: m {s}^{ - 2}}{1600 \: {m}^{2} {s}^{ - 2} }$
$\sin \: \alpha \: = \: \frac{\: 5 \: \times \: 0.7\: \times \: 2 \: \times \: \sqrt{10}}{40}$
$\sin \: \alpha \: = \: \frac{ 7 \sqrt{10} }{40}$
or
$\sin \: \alpha \: = \: 0.5533985905$

Hence,
$\sin \: 2 \alpha \: = \: 0.9218697034$
R = u^2 × sin 2A / g
$r \: = \: \frac{ {(40 \: m {s}^{ - 1} )}^{2} \sin \: 2\alpha }{9.8 \: m {s}^{ - 2} }$
R = 150.5093 m.
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