Answer:
The football would land in [tex]2\; {\rm s}[/tex].
The ball would land [tex]10\; {\rm m}[/tex] away from the base of the wall.
Explanation:
Under the assumptions, the ball would accelerate at a constant [tex]a_{y} = (-g) = (-10)\; {\rm m\cdot s^{-2}}[/tex] in the vertical direction. In this question, the following information are known:
The goal is to find the duration [tex]t[/tex] of the motion. The SUVAT equation [tex]x_{y} = (1/2)\, a_{y}\, t^{2} + u_{y}\, t[/tex] relates these quantities. Since initial velocity [tex]u_{y}[/tex] is zero, this equation becomes:
[tex]x_{y} = (1/2)\, a_{y}\, t^{2}[/tex].
Rearrange this equation to find the duration of the motion, [tex]t[/tex]:
[tex]\begin{aligned}t &= \sqrt{\frac{2\, x_{y}}{a_{y}}} = \sqrt{\frac{2\, (-20\; {\rm m})}{(-10)\; {\rm m\cdot s^{-2}}}} = 2\; {\rm s}\end{aligned}[/tex].
In other words, it would take [tex]2\; {\rm s}[/tex] for the football to land.
Under the assumption that air resistance is negligible, the horizontal velocity of the ball would be constant. Since the initial horizontal velocity is [tex]5\; {\rm m\cdot s^{-1}}[/tex], the velocity of the ball would be [tex]5\; {\rm m\cdot s^{-1}}\![/tex] during the entire motion. To find the distance travelled in the horizontal direction, multiply the horizontal velocity by the duration of the motion:
[tex]\begin{aligned}(5\; {\rm m\cdot s^{-1}})\, (2\; {\rm s}) = 10\; {\rm m}\end{aligned}[/tex].
In other words, the ball would land [tex]10\; {\rm m}[/tex] away from the base of the wall.
