Question

A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height above the ground?

Answer 1

Let us write the expressions that show the dependence of the height on time for every ball

$h_1(t)=0+v_0t-\dfrac{gt^2}{2}, \\ h_2(t)=H+0\cdot t-\dfrac{gt^2}{2}.$

Here $v_0=25\,\text{m/s},\, H=15\,\text{m}.$

When two balls are at the same height, we get

$v_0t-\dfrac{gt^2}{2}=H-\dfrac{gt^2}{2} \text{\;\;\;or} \;\;\; v_0t=H \;\;\Rightarrow \;\; t=\dfrac{H} {v_0}=\dfrac{15\,\text{m}}{25\,\text{m/s}}=0.6\,\text{s}.$