Answer:
33 m/s
Explanation:
By analyzing the vertical motion, we can find what is the time of flight of the projectile. The vertical position is
[tex]y(t) = h + v_{0y}t - \frac{1}{2}gt^2[/tex]
where
h = 20 m is the initial height
[tex]v_{0y} = 26 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
By putting y(t)=0, we find the time t at which the projectile hits the ground:
[tex]0=20 + 26 t - 4.9t^2[/tex]
which has 2 solutions:
t = -0.7 s
t = 6.0 s
We discard the negative solution since it has no physical meaning. So, we know that the projectile hits the ground 6.0 s later after the launch.
The vertical velocity is given by
[tex]v_y (t)= v_{0y} -gt[/tex]
So we can find the vertical velocity when the projectile reaches point Q, by substituting t=6.0 s into this equation:
[tex]v_y = 26 m/s - (9.8 m/s^2)(6.0 s)=-32.8 m/s \sim -33 m/s[/tex]
and the negative sign means the direction is downward.
