# When an object is dropped from a height of 620 m, what is the time taken for it to reach the ground.

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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.

s=u*t+(1/2)*a*(t)^2

s=620

u=0

t=?

a=9.8

t?

620=0t+(1/2)*9.8*t^2

t^2=(620/(1/2*9.8))

t=square root (620/(1/2*9.8))

t=11.24858268

t=11.5 seconds (to 2 d.p)