# Solve this sum

62. The new range will be the total of range and additional distance covered by lorry in time of flight T.

So we need to find out first time of flight and then additional distance covered by lorry in that time.

$RangeR=\frac{{v}^{2}Sin2\theta}{g}\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{Rg}{Sin2\theta}=\frac{360x10}{Sin90}=3600\phantom{\rule{0ex}{0ex}}v=60m/s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}TimeofflightT=\frac{2vSin\theta}{g}=\frac{2\times 60\times Sin45}{10}=8.48s\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Horizontaldis\mathrm{tan}cecoveredbylorryx={v}_{x}T\phantom{\rule{0ex}{0ex}}{v}_{x}=18km/h=5m/s\phantom{\rule{0ex}{0ex}}x={v}_{x}T=5\times 8.48=42.42m\phantom{\rule{0ex}{0ex}}$

So new range will be

360 m +42.42 m = 402.42 m

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