Explanation:
It is given that, a stone travel over level ground if it is thrown upward at an angle of 30.0° with respect to the horizontal.
Speed with which it is thrown, v = 11 m/s
We need to find the maximum height above the ground level. The maximum height attained by the stone is given by :
[tex]h=\dfrac{v^2sin^2\theta}{2g}[/tex]
[tex]h=\dfrac{(11\ m/s)^2sin^2(30)}{2\times 9.8\ m/s^2}[/tex]
h = 1.54 m
So, the stone will travel a distance of 1.54 meters.
Range of a projectile is given by :
[tex]R=\dfrac{v^2sin\ 2\theta}{g}[/tex]
[tex]R=\dfrac{(11\ m/s)^2sin(60)}{9.8\ m/s^2}[/tex]
R = 10.69 m
So, the maximum range achieved with the same initial speed is 10.69 meters. Hence, this is the required solution.
