Question

A boy wants to throw a letter wrapped over a stone to his friend across the street 40m wide the boy's window is 10m below friend's window .how should he throw it ?

Answer 1

A boy wants to throw a letter wrapped over a stone to his friend across the street 40m wide. The boy's window is 10m below friend's window. How should he throw it?

**Solution.**

$$l = 40m, h_{max} = 10m, g = 9.8 \frac{m}{s^2};$$ $$v_0 - ?\alpha - ?$$

A boy should throw a stone so that it will fly in a parabola.

A point "O" is the window of the boy;

A point "A" is the window of the boy's friend.

A Point "A" is the highest point the stone attained.

A boy has to throw a stone at an angle $\alpha$, with an initial speed $v_0$, as in the diagram.

Projections of the speed by the axis:

$$v_x = v_0 \cos \alpha;$$ $$v_y = v_0 \sin \alpha.$$

We obtain the time of the flight of the stone.

The equation of the projection of the speed for the y-axis:

$$v_y = v_{0y} - gt;$$ $$v_y = v_0 \sin \alpha - gt.$$

At the point "A" $v_y = 0$

$$0 = v_0 \sin \alpha - gt;$$ $$v_0 \sin \alpha = gt;$$ $$t = \frac{v_0 \sin \alpha}{g}.$$

The max height attained by the stone.

$$h_{\max} = v_{0y} t - \frac{g t^2}{2};$$ $$h_{\max} = v_0 \sin \alpha t - \frac{g t^2}{2};$$ $$t = \frac{v_0 \sin \alpha}{g}.$$ $$h_{\max} = v_0 \sin \alpha \frac{v_0 \sin \alpha}{g} - \frac{g}{2} \frac{v_0^2 \sin^2 \alpha}{g^2};$$ $$h_{\max} = \frac{v_0^2 \sin^2 \alpha}{g} - \frac{v_0^2 \sin^2 \alpha}{2g};$$ $$h_{\max} = \frac{v_0^2 \sin^2 \alpha}{2g}.$$

The range of the stone:

$$l = v_{0x} t;$$ $$l = v_0 \cos \alpha t;$$ $$l = v_0 \cos \alpha \frac{v_0 \sin \alpha}{g};$$ $$l = \frac{v_0^2 \sin \alpha \cos \alpha}{g}.$$

We have two equations:

1. $h_{max} = \frac{v_0^2\sin^2\alpha}{2g}$

2. $l = \frac{v_0^2\sin\alpha\cos\alpha}{g}$

Divide the second equation by the first and obtain the angle $\alpha$:

$$\frac{l}{h_{max}} = \frac{v_0^2\sin\alpha\cos\alpha 2g}{gv_0^2\sin^2\alpha};$$ $$\frac{l}{h_{max}} = \frac{2\cos\alpha}{\sin\alpha};$$ $$\frac{l}{h_{max}} = \frac{2}{\tan\alpha};$$ $$\tan\alpha = \frac{2h_{max}}{l}.$$ $$\tan\alpha = \frac{2 \cdot 10}{40} = 0.5$$ $$\alpha = 26,565{}^\circ.$$

From the first equation we obtain an initial speed.

$$h_{max} = \frac{v_0^2\sin^2\alpha}{2g};$$ $$v_0 = \frac{\sqrt{2gh_{max}}}{\sin\alpha}.$$ $$v_0 = \frac{\sqrt{2 \cdot 9.8 \cdot 10}}{\sin(26,565{}^\circ)} = 31.3 \left(\frac{m}{s}\right).$$

Answer: A boy should throw a letter wrapped over a stone at an angle $\alpha = 26,565{}^\circ$ at a initial speed of $v_0 = 31.3\frac{m}{s}$.

If the boy is throwing a stone at a lower speed, he will have to increase the angle of throwing.