Answer:
The speed of the projectile when it impacts the ground is 1000 m/s
Explanation:
Vertical Launch
When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.
The speed when the object is moving upwards is given by
[tex]v_f=v_o-g.t[/tex]
The time it takes to reach the maximum height is when vf=0, i.e.
[tex]0=v_o-g.t[/tex]
solving for t
[tex]\displaystyle t=\frac{v_o}{g}[/tex]
The maximum height reached is
[tex]\displaystyle y=\frac{gt^2}{2}=\frac{v_o^2}{2g}[/tex]
Then, the object starts to fall. The object's height is given by
[tex]\displaystyle y=\frac{v_o^2}{2g}-\frac{gt'^2}{2}[/tex]
where t' is the time the object has traveled downwards. The height will be 0 again when
[tex]\displaystyle \frac{v_o^2}{2g}=\frac{gt'^2}{2}[/tex]
Solving for t'
[tex]\displaystyle t'=\frac{v_o}{g}[/tex]
We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is
[tex]v_f=g.t'=v_o[/tex]
Thus, the speed of the projectile when it impacts the ground is 1000 m/s
