Question

Can someone help me?!!!!!

Answer 1

33 m/s

Step-by-step 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 - (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.