Answer to Question #134578 in Physics for Rancis B. Wood

Question

A ball is batted into the air and caught at a point 100 m distance horizontally in 4.0 seconds. Neglecting air resistance. Find the maximum height attain by the ball.

Answer 1

The height of a body thrown at some angle above horizontal:

$$h=\frac{v^2\text{sin}^2\theta}{2g}.$$

The range is

$$R=\frac{v^2\text{ sin}(2\theta)}{g},\\\space\\ \text{ sin}(2\theta)=2\text{cos}\theta\text{ sin}\theta,\\\space\\ R=\frac{2v^2\text{cos}\theta\text{ sin}\theta}{g}.$$

On the other hand, the range is

$$R=v\text{ cos}\theta \cdot t,\\\space\\ v\text{ cos}\theta =\frac{R}{t}.$$

Substitute to the equation with sine of a double angle:

$$R=\frac{2Rv\text{ sin}\theta}{gt},\\\space\\ v\text{ sin}\theta=\frac{gt}{2}.$$

Substitute this into the first equation for height:

$$h=\frac{(gt/2)^2}{2g}=\frac{gt^2}{8}=19.6\text{ m}.$$