Question

Can someone help me?!!!!!

Answer 1

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

$y(t) = h + v_{0y}t - \frac{1}{2}gt^2$

where

h = 20 m is the initial height

$v_{0y} = 26 m/s$ is the initial vertical velocity

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

By putting y(t)=0, we find the time t at which the projectile hits the ground:

$0=20 + 26 t - 4.9t^2$

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

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

So we can find the vertical velocity when the projectile reaches point Q, by substituting t=6.0 s into this equation:

$v_y = 26 m/s - (9.8 m/s^2)(6.0 s)=-32.8 m/s \sim -33 m/s$

and the negative sign means the direction is downward.