Answer:
56 m/s
Explanation:
The time we are considering is
t = 15 s
The vertical velocity of the projectile is given by
[tex]v_y(t) = v_{0y}-gt[/tex]
where
[tex]v_{0y}=100 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Substituting t=15 s, we find the vertical velocity of the projectile at that time:
[tex]v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s[/tex]
where the negative sign means the direction is now downward.
The horizontal velocity does not change since there are no forces acting along that direction, so it remains constant:
[tex]v_x = 30 m/s[/tex]
So, the magnitude of the velocity at the moment of impact is
[tex]v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s[/tex]
