Answer :
Answer and explanation:
Let [tex]v_1[/tex] be the velocity of a body projected vertically up, 1 second before it reaches its maximum height. We know, from the kinematics equations, that the distance traveled [tex]y[/tex] in an interval of time [tex]t[/tex] is equal to:
[tex]y=v_0t-\frac{1}{2}gt^{2}[/tex]
Then, in the last second, the distance traveled [tex]y_1[/tex] is equal to:
[tex]y_1=v_1t-\frac{1}{2}gt^{2}[/tex]
But the velocity [tex]v_1[/tex] is related to the time by the equation:
[tex]t=\frac{v_1}{g}\\\\v_1=gt[/tex]
And substituting this expression in the equation above, we obtain:
[tex]y_1=gt^{2}-\frac{1}{2}gt^{2}\\\\y_1=\frac{1}{2}gt^{2}\\\\y_1=\frac{1}{2}(9.8m/s^{2})(1s)^{2}\\ \\y_1=4.9m[/tex]
It means that the distance traveled by the body in the last second of its motion is a constant, independent of its initial velocity.