Answer:
[tex]t=12.25\ seconds[/tex]
Explanation:
Diagonal Launch
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
[tex]x=v_ocos\theta t[/tex]
[tex]\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}[/tex]
Where vo is the magnitude of the initial velocity, [tex]\theta[/tex] is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
[tex]\displaystyle t=\frac{v_osin\theta}{g}[/tex]
There are two times where the value of y is [tex]y_o[/tex] when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making [tex]y=y_o[/tex]
[tex]\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}[/tex]
Removing [tex]y_o[/tex] and dividing by t (t different of zero)
[tex]\displaystyle 0=v_osin\theta-\frac{gt}{2}[/tex]
Then we find the total flight as
[tex]\displaystyle t=\frac{2v_osin\theta}{g}[/tex]
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
[tex]\boxed{t=24.5/2=12.25\ seconds}[/tex]
