Question

The figure shows top view of a boat moving with velocity $\vec{v}_{r}=(8 \hat{i}+6 \hat{j}) m / s$ relative to water. Water flows with velocity $\vec{v}_{w}=2 \hat{i} \,m / s$ relative to ground. A motorcyclist moves along a road running parallel to bank with velocity $\vec{v}_{m}=20 \hat{i}\, m / s$. An apple thrown from the boat is caught by the motorcyclist in same horizontal plane after $4\, s$ of projection. If apple is thrown with velocity $v_{x} \hat{i}+v_{y} \hat{j}+v_{z} \hat{k}$ relative to boat, then find the value of $\frac{\left|\vec{v}_{x}\right|+\left|\vec{v}_{y}\right|+\left|\vec{v}_{z}\right|}{13}$.

Answer 1

$v_{b(\text { boat })}=10 \hat{i}+6 \hat{j}$
$v_{a \text { (apple) }}=\left(v_{x}+10\right) \hat{i}+\left(v_{y}+6\right) \hat{j}+v_{z} \hat{k}$
$v_{\text {apple wrt. cyclist }}=v_{a c}=\left(v_{x}-10\right) \hat{i}+\left(v_{y}+6\right) \hat{j}+v_{z} \hat{k}$
$\Rightarrow v_{x}-10=0$
$\Rightarrow v_{x}=10 \,m / s$
Now, range $=60=\left(v_{y}+6\right) \times 4$
$\Rightarrow v_{y}=9\,m / s $
$\Rightarrow T=4=\frac{2 v_{z}}{g}$
$ \Rightarrow v_{z}=20 \,m / s$
Thus, $\frac{\left|v_{x}\right|+\left|v_{y}\right|+\left|v_{z}\right|}{13}$
$=\frac{10+9+20}{13}=3$