# A body is thrown up with a speed of 49m/s. It travels 5m in the last second of its upward journey. If the same body is thrown up with a velocity of 98m/s, how much distance (in m) will it travel in the last second of its upward journey? We'll ignore air resistance. Denote the gravitational acceleration as `g ((m)/(s^2)).`

Then the speed is `V(t) = V_0 - g t,` where `V_0` is the initial upward speed. The last moment of an upward journey is that moment when `V(t) = 0.` Therefore `1` second before this the speed (denote it as `V_1`) is greater than zero by `g*1 m/s.` It is `g m/s` regardless of `V_0.`

The distance traveled during the last second is `dH = V_1 t-(g t^2)/2,` and recall that `t = 1 s` and `V_1 = g m/s.` Thus `dH = g - g/2 = g/2` regardless of the initial speed.

So the answer is: the distance is 5 m, the same as in the first case.

Note 1. The result of `5 m` means that the value of `g = 10 m/s^2` was used.

Note 2. For initial speeds less than `10 (m)/s` the above reasoning appears incorrect.

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