Question

An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one

Answer 1

Answer:

20.78 m/s that we can approximate to option d (21 m/s)

Explanation:

The solution involves a lot of algebra and to be familiar with different convenient formulas for launching an object vertically under the action of gravity.

First you need to recall (or derive) the formula for the maximum height reached by an object with launches with initial velocity $v_0$:

Maximum height = $\frac{{v_0}^2}{2g}$

Therefore one fourth of such height would be: $\frac{{v_0}^2}{8g}$

Second, find what would be the time needed to reach that height by solving for the time in the equation for the vertical position:

$y(t)=v_0*t-\frac{g}{2} t^2 \\\frac{{v_0}^2}{8g} = v_0*t-\frac{g}{2} t^2\\\frac{g}{2} t^2-v_0*t+\frac{{v_0}^2}{8g}=0$

And now, solve for t in the last equation using the quadratic formula to find the time needed for the object to reach that height (one fourth of the max height):

$t=\frac{v_0+/-\sqrt{{v_o}^2-4*\frac{g}{2}*\frac{{v_o}^2}{8g} } }{g} = \frac{v_0+/-\sqrt{{v_o}^2-\frac{{v_o}^2}{4} } }{g} =\frac{v_0+/-\sqrt{\frac{3{v_o}^2}{4} } }{g} = \\=\frac{v_0+/-v_0\sqrt{\frac{3}{4} } }{g} =\frac{v_0+/-v_0\frac{\sqrt{3}}{2} } {g}$

Next, use this expression for t in the equation for the velocity at any time t in the object's trajectory that comes from the definition of acceleration;

$v(t)=v_0-g*t$

Then for the time we just found, this new equation becomes:

$v=v_0-g(\frac{v_0+/-v_0\frac{\sqrt{3} }{2}}{g}) =v_0-v_0+/- v_0\frac{\sqrt{3} }{2} = +/- v_0 \frac{\sqrt{3} }{2}$

Now, using that the velocity at this height is 18 m/s, and solving for the unknown velocity $v_0$, we get:

$v_0=\frac{18*2}{\sqrt{3} } =\frac{36}{\sqrt{3} }= 20.78\frac{m}{s}$