46. A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.
First, we must know how long it takes to the first ball to get to h/2.
\(y_f=h/2\\y_o=h/2\\a=-9.8m/s^2=-g\\v_o=0m/s\\t=?\)
\(y_f=y_o+v_ot+\frac{at^2}{2}\)
\(h/2=h+\frac{(-g)t^2}{2}\)
\(h/2=h-\frac{gt^2}{2}\)
\(-h/2+h=\frac{gt^2}{2}\\ h/2=\frac{gt^2}{2}\\t=\sqrt{\frac{h}{g}}\)
As both balls meet at a height h/2 then the second ball meet the first one at the same time. So we just need to substitute the value of time found for the ball dropped into the equation of motion of the second one and later just solve for \(v_o\) the initial velocity necessary so that they can meet at that distance.
For the second ball;
\(y_f=h/2\\y_o=0m\\a=-9.8m/s^2=-g\\v_o=?\)
First, we must know how long it takes to the first ball to get to h/2.
\(y_f=h/2\\y_o=h/2\\a=-9.8m/s^2=-g\\v_o=0m/s\\t=?\)
\(h/2=h+\frac{(-g)t^2}{2}\)
\(h/2=h-\frac{gt^2}{2}\)
\(-h/2+h=\frac{gt^2}{2}\\ h/2=\frac{gt^2}{2}\\t=\sqrt{\frac{h}{g}}\)
As both balls meet at a height h/2 then the second ball meet the first one at the same time. So we just need to substitute the value of time found for the ball dropped into the equation of motion of the second one and later just solve for \(v_o\) the initial velocity necessary so that they can meet at that distance.
For the second ball;
\(y_f=h/2\\y_o=0m\\a=-9.8m/s^2=-g\\v_o=?\)
\(h/2=v_o(\sqrt{\frac{h}{g}})-\frac{g(\sqrt{\frac{h}{g}})^2}{2}\\\)
\(h/2=v_o(\sqrt{\frac{h}{g}})-h/2\)
\(h/2+h/2=v_o(\sqrt{\frac{h}{g}})\\ v_o=\sqrt{\frac{g}{h}}*h=\boxed{\sqrt{gh}}\)
\(h/2+h/2=v_o(\sqrt{\frac{h}{g}})\\ v_o=\sqrt{\frac{g}{h}}*h=\boxed{\sqrt{gh}}\)