Please explain this. will only give points to a fully explained answer Some rela

Question

Please explain this. will only give points to a fully explained answer

Some relations applying to uniformly accelerated motion are: v=v0 +at, V^2=v0^2 +2a(x?x0), and x= x0 +vot+1/2 at^2 At a secret lab a railgun for launching s e probes is under development. To launch a probe vertically from sea level into space the railgun must accelerate the probe to a velocity of 21 km s . (i) [(the railgun can apply a constant acceleration of 17 km s^2 then how long does the railgun have to be to achieve the correct launch velocity for the probe? (ii) Once the probe leaves the railgun it will experience a constant average acceleration of -500 ms^-2 due to air resistance. Assuming the probe's path is vertical and Earth's atmosphere is 150 km thick, calculate the probe's velocity as it leaves the atmosphere.

Explanation / Answer

i)given the acceleration a = 17000 m/s^2

u = 0 m/sec

v = 21000 m/sec

we use v^2 - u^2 = 2as

s = (v^2 - u^2)/2a

s = (21000^2)/(2*17000) = 12970 m = 12.97 Km

2) here given S = 150000 m

a = -500 m/s^2

u = 21000 m/sec

using v^2 - u^2 = 2as

v = (u^2+2as)^0.5

By substituting all the values

v = ((21000^2) -(2*500*150000))^0.5 =17058 m/sec = 17.058 Km/sec

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