Answer:
a) [tex]t=\sqrt{\frac{2h-2x*tan \ \theta}{g} }[/tex]
b)[tex] v_0=\frac{x}{tsin \theta}[/tex]
c)[tex]v_f=\sqrt{v_x^2+v_y^2}[/tex]
[tex]tan \theta=\frac{v_y}{v_x}[/tex]
Explanation:
Since you are not giving the height I will state it as h, then you can just replace and find the numerical value.
The problem can be divided in vertical and horizontal components, which are independant.
To find initial velocity in both components just find them using trigonometry:
[tex]v_{0y} = v_0sin \ \theta\\v_{0x} = v_0cos \ \theta[/tex]
[tex]h=v_{0y}t+\frac{1}{2} gt^2[/tex]
for horizontal flight use:
[tex]v_{0x}=\frac{x}{t}[/tex]
where x=57m
From these two equations and using the decomposition of the initial velocity you can solve for t:
[tex]t=\sqrt{\frac{2h-2x*tan \ \theta}{g} }[/tex]
once you know the time, you can find the initial velocity:
[tex]v_{0x} = \frac{x}{t}=v_0sin \theta\\ v_0=\frac{x}{tsin \theta}[/tex]
Then, you need to calculate the final velocityin both components.
In the horizontal, the velocity does not change because there is no force in this direction, while in the vertical direction:
[tex]2gh=v_{fy}^2-v_{0y}^2\\v_{fy}=\sqrt{2gh-v_{0y}^2}\\v_{fy}=\sqrt{2gh-(v_0sin \theta)^2}\\[/tex]
To find the speed just use this formula:
[tex]v_f=\sqrt{v_x^2+v_y^2}[/tex]
and to find the angle use:
[tex]tan \theta=\frac{v_y}{v_x}[/tex]
