Answer to Question #145252 in Physics for XYZ

Question

An Object is thrown into space with a velocity of 15 m/s at an angle of 60o with
respect to the horizontal, Calculate (i) maximum height
(ii) time of flight

(iii) horizontal range. (Given g = 9.8 m/s-2
)

Answer 1

See https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/ for more details.

1. The maximum height for a body launched at an angle to the horizontal, is:

$$h_{max} = \dfrac{v^2\sin^2\theta}{2g}$$

where $v=15m/s$ is the initial speed of the body, $\theta = 60\degree$ is launching angle and $g = 9.8m/s^2$ is the gravitational acceleration. Thus, obtain:

$$h_{max} = \dfrac{15^2\sin^260\degree}{2\cdot 9.8} \approx 8.60m$$

2. The time of flight is:

$$T = \dfrac{2v\sin\theta}{g} = \dfrac{2\cdot 15\sin60\degree}{9.8} \approx 2.65 s$$

3. The horizontal range:

$$R = \dfrac{v^2\sin2\theta}{g} = \dfrac{15^2\sin(2\cdot 60\degree)}{9.8} \approx 19.9m$$

Answer. i) 8.60m, ii) 2.65s, iii) 19.88m.