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The distance traveled by an object (d) moving with an initial speed (u) and which is accelerating at a rate a is t where d = u*t + (1/2)*a*t^2.
In the problem, the object has been dropped from a height of 620 m, the initial speed of the object in this case is 0 m/s. The gravitational force of attraction accelerates the object downwards at 9.8 m/s^2. The distance traveled by the object before it strikes the ground is 620 m.
Substituting d = 620, a = 9.8, u = 0 gives the equation:
`620 = 0*t + (1/2)*9.8*t^2`
=> `620 = (1/2)*9.8*t^2`
=> `t^2 = 6200/49`
=> `t ~~ 11.24`
When dropped from a height of 620 m, the object strikes the ground after 11.24 seconds.
t=square root (620/(1/2*9.8))
t=11.5 seconds (to 2 d.p)
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