Answer to Question #149211 in Physics for Maitha

Question

A ball is launched at 4.5 m/s at 66° above the horizontal. It starts and lands at the same distance from the ground. What are the maximum height above its launch level and the flight time of the ball?

Answer 1

1. Time, required to reach the highest point is (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$t = \dfrac{2v_0\sin\theta}{g}$$

where $v_0 = 4.5m/s$ is the initial speed, $\theta = 66\degree$ is the launch angle, and $g = 9.81m/s^2$ is the gravitational acceleration. Thus, obtain:

$$t = \dfrac{2\cdot 4.5\cdot \sin66\degree}{9.81} \approx 0.84\space s$$

2. The maximum height above its launch level is:

$$H = \dfrac{v_0^2\sin^2\theta}{2g} = \dfrac{4.5^2\sin^266\degree}{2\cdot 9.81} \approx 0.86m$$

Answer. The maximum height above its launch level is 0.86 m and The flight time of the ball is 0.84 s