Answer
a) The mat has to be placed $55.35 m$ south of the point where the stuntwoman drops from the helicopter.
b) Graphs are shown in Figure 3.48.
Work Step by Step
The given problem is an example of projectile motion where the stuntwoman falls under the action of gravitational force from the helicopter.
Let us take the +y direction as upward.
In the question, it is given that the helicopter is at a height of $30m$ above the ground and moves with a constant velocity, The vertical and horizontal components of this velocity are $10 m/s$ and $15 m/s$ respectively. Furthermore, 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 stuntwoman. Therefore, the horizontal acceleration ($a_{x}$) equals zero. Since the stuntwoman 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) The horizontal distance covered by the stuntwoman is determined by the time she spent in the air during her vertical fall.
The initial velocity with which she made the jump is equal to the constant velocity components of the helicopter.
Therefore,
$u_{y}=10m/s$
$u_{x}=15m/s$
The time taken by the stuntwoman to cover the 30 m height from the helicopter in the air can be found by applying the second equation of motion in the vertical direction.
$s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2}$
$s_{y}=u_{y}t-\frac{1}{2}gt^{2}$
substituting the known values, we get
$-30=10t-\frac{1}{2}\times9.8t^{2}$ (minus sign before 30 is because, here the distance is measured from the helicopter to the ground in the -y direction)
$-30=10t-4.9t^{2}$
on rearranging we get
$4.9t^{2}-10t-30=0$
which is a quadratic formula in $t$ having a form $At^{2}+Bt+c=0$ . This quadratic formula can be solved for $t$ using the equation, $t=\frac{-B\pm \sqrt (B^{2}-4AC)}{2A}$ .
Therefore,
$t=\frac{-(-10)\pm\sqrt ((-10)^{2}-4\times4.9\times(-30))}{2\times4.9}$
$t=\frac{10\pm\sqrt (100+588)}{9.8}=\frac{10\pm\sqrt (688)}{9.8}=\frac{10\pm26.22}{9.8}$
which implies
$t=\frac{10+26.22}{9.8}=3.69s$ or $t=\frac{10-26.22}{9.8}=-1.65s$
Since time cannot be negative, the time taken by the stuntwoman is $3.69s$.
The horizontal distance covered by the stuntwoman at this time can be found using the second equation of motion in the horizontal direction.
$s_{x}=u_{x}t+\frac{1}{2}a_{x}t^{2}$
since $a_{x}=0$
$s_{x}=u_{x}t$
$s_{x}=15\times3.69=55.35m$
Thus, the stuntwoman will land $55.35 m$ south of the point where she drops from the helicopter and this is where the mats should have been placed to break her fall.
b) Graphs are shown in Figure 3.48.
