# An object is thrown straight up from ground level with a speed of 50 m/s. If g = 10 m.s^2, what is its distance above ground level 1.0 s later.

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The distance traveled by an object in t seconds if its starts at an initial velocity of u m/s and accelerates at a m/s^2 is given by `s = u*t + (1/2)*a*t^2` .

Here u = 50 m/s, a = -10 m/s and t = 1

Substituting these values in the formula for distance traveled:

s = 50*1 - (1/2)*10*1 = 50 - 5 = 45 m

**The object is at a distance above the ground equal to 45 m after 1 second.**