Question

Can someone help me?!!!!!

Answer 1

56 m/s

Step-by-step 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=√(v_x^2 +v_y^2)=√((30 m/s)^2+(-47 m/s)^2)=55.8 m/s \sim 56 m/s`