Question

A cannon and a target are 5100m apart and located at the same level. How soon will the shell launched with the initial velocity 240m/s reach the target? Plz answer it..... I need it immediately

Answer 1

Distance = 5100m
Initial velocity = 240 m/s
Hence it will come to rest on hitting the target therefore it's final velocity will be 0 m/s.
We will apply the equation of motion which is
${v}^{2} = {u}^{2} + 2as \\ 0 = {240}^{2} + 2 \times 5100a \\ - 57600 = 10200a \\ a = - \frac{57600 }{10200} \\ - \frac{576}{102} = a \\ a = - \frac{192}{34} \\ a = - \frac{96}{17} m {s}^{ - 2}$
And then we will apply v = u+at
0 = 240 m/s - 96/17 t
-240 m/s = - 96 / 17 t
$t = \frac{240 \times 17}{96}s \\ t = 42.5 \: s$
Hence t = 42.5 seconds
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