A rock can be launched toward a wall from ground level with initial velocity of v = (3i+4j)m/s. The distance between the launch site and the wall is 15 meters. What are the least speed an the greatest speed with which the rock can hit the wall?

When an object is launched with an initial velocity that has both horizontal and vertical components, it describes a parabola in the air and eventually returns to the ground after reaching its maximum height.

We have the initial velocity as

[tex]v_o=(3\hat i+4\hat j)\ m/s[/tex]

And we also know the launching site is 15 meters away from the wall. That's all we need to know when the rock reaches the wall (if it does). The horizontal distance is

[tex]x=v_{ox}.t[/tex]

Where vox is the horizontal component of the initial velocity. Solving for t

[tex]\displaystyle t=\frac{x}{v_{ox}}[/tex]

[tex]\displaystyle t=\frac{15}{3}=5\ sec[/tex]

We'll show this is an absurd time given the conditions of the problem. The height of the rock is computed by

[tex]\displaystyle y=v_{oy}.t-\frac{gt^2}{2}[/tex]

If we use the value of t previously found

[tex]\displasystyle y=4\times 5-\frac{9.8\times 5^2}{2}[/tex]

[tex]y=-102.5\ m[/tex]

This means the rock will never reach the wall since it lands on the ground way before that. The problem has no solution