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Dima020 [189]
3 years ago
8

Can someone help me?!!!!!

Physics
1 answer:
babunello [35]3 years ago
3 0

Answer:

56 m/s

Explanation:

The time we are considering is

t = 15 s

The vertical velocity of the projectile is given by

v_y(t) = v_{0y}-gt

where

v_{0y}=100 m/s is the initial vertical velocity

g=9.8 m/s^2 is the acceleration due to gravity

Substituting t=15 s, we find the vertical velocity of the projectile at that time:

v_y = 100 m/s - (9.8 m/s^2)(15 s)=-47 m/s

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:

v_x = 30 m/s

So, the magnitude of the velocity at the moment of impact is

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

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