Question

Show that a body projected vertically up from the ground, the distance travelled by it in the last second of its upwards motion is an constant independent of its initial velocity

Answer 1

Answer and explanation:

Let $v_1$ 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 $y$ in an interval of time $t$ is equal to:

$y=v_0t-\frac{1}{2}gt^{2}$

Then, in the last second, the distance traveled $y_1$ is equal to:

$y_1=v_1t-\frac{1}{2}gt^{2}$

But the velocity $v_1$ is related to the time by the equation:

$t=\frac{v_1}{g}\\\\v_1=gt$

And substituting this expression in the equation above, we obtain:

$y_1=gt^{2}-\frac{1}{2}gt^{2}\\\\y_1=\frac{1}{2}gt^{2}\\\\y_1=\frac{1}{2}(9.8m/s^{2})(1s)^{2}\\ \\y_1=4.9m$

It means that the distance traveled by the body in the last second of its motion is a constant, independent of its initial velocity.