Question

a baseball is thrown upwards and leaves the hand at a height of 1.6 above the level playing field. the ball has an initial speed of 26 m/sec at an angle of 24 above the horizontal. determine the maximum height reached and the range

Answer 1

The maximum height:

$$H=h+\frac{(v\sin{24\degree})^2}{2g}$$

$$H=1.6+\frac{(26\sin{24\degree})^2}{2(9.8)}=7.3\ m$$

The range:

$$R=\frac{v^2}{2g}\left(1+\sqrt{1+\frac{2gh}{(v\sin{24\degree})^2}}\right)\sin{(2(24\degree))}$$

$$R=\frac{26^2}{2(9.8)}\left(1+\sqrt{1+\frac{2(9.8)(1.6)}{(26\sin{24\degree})^2}}\right)\sin{(48\degree)}=55\ m$$