Answer
a) Time required to reach the highest point of the trajectory$= 1.22 s$.
b) Highest point in the trajectory $=7.35 m$.
c) Time required to return to the original level $=2.45s$ , which is twice that found in part (a).
d) Horizontal distance travelled by the football$=49m$.
e) The graphs are drawn in Figure 3.12.
Work Step by Step
The given problem is an example of projectile motion where the football falls under the action of gravitational force.
Let us take the +y direction as upward.
In the question, it is given that the initial velocity of the football has both vertical ($u_{y}$) and horizontal ($u_{x}$) components with magnitude
$u_{y}=12 m/s$
$u_{x}=20 m/s$
In the question, it is asked to ignore the air resistance which implies that there is no horizontal force (frictional force due to air drag) acting on the football. Therefore, the horizontal acceleration ($a_{x}$) equals zero. Since the football falls under gravity, the vertical component of acceleration is $a_{y}=-g$ (the negative sign implies the acceleration is acting in the downward direction).
a) In projectile motion, the vertical velocity component at the highest point is zero.
therefore, by applying the first equation of motion in the vertical direction we get,
$v_{y}=u_{y}+a_{y}t$
$v_{y}=u_{y}-gt$
since $v_{y}=0$
$gt=u_{y}$
$ t=\frac{u_{y}}{g}$
$ t=\frac{12}{9.8}=1.22 s$
b) To find the maximum height attained by the football we can apply the third equation of motion in the vertical direction
$(v_{y})^{2}=(u_{y})^{2}+2ah$
where $h$ is the maximum vertical displacement.
$(v_{y})^{2}=(u_{y})^{2}-2gh$
since $v_{y}=0$
$2gh=(u_{y})^{2}$
$h=\frac{(u_{y})^{2}}{2g}$
$h=\frac{12^{2}}{2\times9.8}=7.35 m$
c) Let $t$ be the total time taken by the football. As football returns to its original level, the net displacement in the vertical direction ($s_{y}$) will be zero. Therefore applying the second equation of motion in the vertical direction,
$s_{y}=u_{y}t+\frac{1}{2}at^{2}$
$s_{y}=u_{y}t-\frac{1}{2}gt^{2}$
since $s_{y}=0$
$\frac{1}{2}gt^{2}-u_{y}t=0$
factorising for $t$, we get
$t(\frac{1}{2}gt-u_{y})=0$
which implies
either $t=0$, representing the initial condition
or
$\frac{1}{2}gt-u_{y}=0$
$t=\frac{2u_{y}}{g}=\frac{2\times12}{9.8}=2.45s$, for the total flight which is twice that found in part (a).
d) By considering the horizontal component of velocity and the total time taken for the entire flight, we can determine the horizontal distance traveled by the football.
Horizontal distance $H=u_{x}t$
$H=20\times2.45=49m$
e) The graphs are drawn in Figure 3.12 using the equations of motion
