Answer to Question #306986 in Physics for Jen

Question

Suppose a golf ball is hit off with the tee with an initial velocity of 30 m/s at an angle of 35° to the horizontal.

What is the maximum height reached by the ball?

What is its range?

Answer 1

The maximum height is given as follows (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

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

where $u = 30m/s, \theta = 35\degree , g = 9.8m/s^2$. Thus, obtain:

$$h = \dfrac{30^2\cdot \sin35\degree}{2\cdot 9.8} \approx 26m$$

The range is:

$$R = \dfrac{u^2\sin2\theta}{g} = \dfrac{30^2\sin2\cdot 35\degree}{9.8} \approx 86m$$