Question

Can someone help me?!!!!!

Answer 1

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

$v=\sqrt{v_x^2 +v_y^2}=\sqrt{(30 m/s)^2+(-47 m/s)^2}=55.8 m/s \sim 56 m/s$