Answer to Question #293646 in Physics for Exes

Question

A baseball is shut with an initial speed of 12.0m/s.

(a) how far does it go in the horizontal direction if it is shut at angle (I) 15° (ii) 50° (III) 45 and (iv) 60° above the horizontal

(B) what is the maximum height reached in the 4 cases?

Answer 1

The range and height are:

$$R=\dfrac{v^2\sin(2\theta)}{g},\\\space\\ H=\dfrac{(v\sin\theta)^2}{2g}.$$

The results below are the range and height respectively:

(I) 15°: 7.3 m. 0.49 m.

(ii) 50°: 14.5 m. 4.31 m.

(III) 45°: 14.7 m. 3.67 m.

(iv) 60°: 12.7 m. 5.51 m.