Solution for Question 12
The grasshopper performs projectile motion. Let the origin be the launch point. The position of the grasshopper $(x_g, y_g)$ at any time $t$ is given by:
- Horizontal position: $x_g(t) = u \cos \theta \cdot t$
- Vertical position: $y_g(t) = u \sin \theta \cdot t – \frac{1}{2} g t^2$
The sun rays make an angle $\phi$ with the horizontal. Since the grasshopper jumps “towards the sun”, the sun is positioned in the direction of motion. This geometry means the shadow $(x_s)$ falls behind the grasshopper’s horizontal position ($x_g$).
Consider the triangle formed by the grasshopper, the shadow, and the ground projection. From the diagram:
$$ \tan \phi = \frac{y_g – 0}{x_g – x_s} $$
Rearranging to solve for the shadow’s position $x_s$:
$$ x_g – x_s = \frac{y_g}{\tan \phi} = y_g \cot \phi $$
$$ x_s = x_g – y_g \cot \phi $$
To find the velocity of the shadow, we differentiate its position with respect to time:
$$ v_s = \frac{dx_s}{dt} = \frac{dx_g}{dt} – \frac{dy_g}{dt} \cot \phi $$
Substituting the known velocity components of the grasshopper:
$$ \frac{dx_g}{dt} = v_x = u \cos \theta $$
$$ \frac{dy_g}{dt} = v_y = u \sin \theta – gt $$
Now, substitute these into the shadow velocity equation:
$$ v_s = u \cos \theta – (u \sin \theta – gt) \cot \phi $$
$$ v_s = u \cos \theta – u \sin \theta \cot \phi + gt \cot \phi $$
Factor out $u$ and use trigonometry identities ($\cot \phi = \frac{\cos \phi}{\sin \phi}$):
$$ v_s = u \left(\cos \theta – \sin \theta \frac{\cos \phi}{\sin \phi}\right) + gt \cot \phi $$
$$ v_s = u \left(\frac{\sin \phi \cos \theta – \cos \phi \sin \theta}{\sin \phi}\right) + gt \cot \phi $$
Using the subtraction formula $\sin(A – B) = \sin A \cos B – \cos A \sin B$: