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 $4s$ 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 $\frac{|{\vec{v}}_x|+|{\vec{v}}_y|+|{\vec{v}}_z|}{13}$.

Answer 1

Answer: 3

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