The time of flight of the arrow is 14.96 s
Explanation:
The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
In order to find the time of flight of the arrow, we just need to analyze its vertical motion. We can do it by using the following suvat equation:
[tex]v_y = u_y + at[/tex] (1)
where
[tex]v_y[/tex] is the vertical velocity at time t
[tex]u_y[/tex] is the initial vertical velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
The arrow reaches its highest point in the trajectory when the vertical velocity, so when
[tex]v_y=0[/tex]
Also, the initial vertical velocity is given by
[tex]u_y = u sin \theta = (83.0)(sin 62.0^{\circ})=73.3 m/s[/tex]
Therefore, from (1) we find
[tex]t=-\frac{u_y}{g}=-\frac{73.3}{-9.8}=7.48 s[/tex]
This is the time the arrow needs to reach the highest point: the total time of flight of the arrow is just twice this time, so
[tex]T=2t=14.96 s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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