Question

a ball is projected from the ground at aspeed of 10 m/s at an angle of 30 with horizontal . another ball is simultaneously released from a point on a vertical line along the maximum height of the projectile . boyh yhe projectile collide at the maximum height of the projectile . what was the initial height of the second ball

Answer 1

$$\begin{gathered} maximumheightoftheprojectile=h=\frac{u^2{\sin }^2\theta }{2g}=\frac{100\times {\displaystyle \frac{1}{4}}}{2\times 10}=10/8m \\ tisthetimeafterwhichtheycollide. \\ 0=u\sin \theta -gt \\ t=10\times \frac{1}{2\times 10}=0.5s \\ dis\tan cetravelledbythesecondballintimet, \\ H=0+\frac{1}{2}g{t}^2=0.5\times 10\times 0.5\times 0.5=1.25m \\ totalheightofthesecondball;=1.25+1.25=2.5m \end{gathered}$$