Answer:
The height from which the rock was thrown is 1.92 m
Solution:
As per the question:
Speed with which the rock is thrown, v = 12.0 m/s
Horizontal distance traveled by the rock before it hits the ground, d = 15.5 m
Launch angle, ![\theta = 30.0^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2030.0%5E%7B%5Ccirc%7D)
Now,
To calculate the height, h from which the rock was thrown:
First, since we consider the horizontal motion in the trajectory of the rock, thus the time taken is given by:
![t = \frac{d}{vcos\theta}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bd%7D%7Bvcos%5Ctheta%7D)
![t = \frac{15.5}{12.0cos30.0^{\circ}} = 1.49\ s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B15.5%7D%7B12.0cos30.0%5E%7B%5Ccirc%7D%7D%20%3D%201.49%5C%20s)
Now,
The height from which the rock was thrown is given by the kinematic eqn, acceleration in the horizontal direction is zero:
![h = vsin\theta t - \frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=h%20%3D%20vsin%5Ctheta%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D)
![h = 12.0sin30.0^{\circ}\times 1.49 - \frac{1}{2}\times 9.8\times 1.49^{2} = - 1.92\ m](https://tex.z-dn.net/?f=h%20%3D%2012.0sin30.0%5E%7B%5Ccirc%7D%5Ctimes%201.49%20-%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%201.49%5E%7B2%7D%20%3D%20-%201.92%5C%20m)