Answer:
t=14.96 sec
Explanation:
Diagonal Launch
It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.
The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is
:
[tex]\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}[/tex]
Where vo is the initial speed, [tex]\theta[/tex] is the angle, t is the time and g is the acceleration of gravity
.
In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus [tex]y_o=0[/tex], and:
[tex]\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}[/tex]
The value of y is zero twice: when t=0 (at launching time) and in t=[tex]t_f[/tex] when it goes back to the ground. We need to find that time [tex]t_f[/tex] by making [tex]y=0[/tex]
[tex]\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}[/tex]
Dividing by [tex]t_f[/tex]
[tex]\displaystyle v_osin\theta=\frac{gt_f}{2}[/tex]
Then we find the total flight time as
[tex]\displaystyle t_f=\frac{2v_osin\theta}{g}[/tex]
[tex]\displaystyle t_f=\frac{2(83)sin\ 62^o}{9.8}[/tex]
[tex]\displaystyle t_f=14.96\ sec[/tex]
