Answer to Question #249128 in Mechanics | Relativity for mhia

Question

A ball is batted into the air and the ball found at a point 200 m distance horizontally in

5.0 seconds. Neglecting air resistance. Find the maximum height attain by the ball.

Answer 1

The range $R=200m$ and the time in air $T = 5s$ are given as follows (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$R = \dfrac{u^2\sin2\theta}{g} \\ T = \dfrac{2u\sin\theta}{g}$$

where $u$ is the initial speed, $\theta$ is the launch angle, $g = 9.81m/s^2$ is the gravitational acceleration.

The maximum height is given as follows:

$$h = \dfrac{(u\sin\theta)^2}{2g}$$

Expressing from the formula for time quantity $u\sin\theta$ and substituting it into the expression for $h$, obtain:

$$u\sin\theta = \dfrac{gT}{2}\\ h = \dfrac{g^2T^2}{4\cdot 2g} = \dfrac{gT^2}{8}\\ h = \dfrac{9.81m/s^2\cdot (5s)^2}{8} \approx 31m$$

Answer. 31 m.