Question

A body of mass m is thrown with velocity υ at angle of 300 to the horizontal and
another body B of the same mass is thrown with velocity υ at an angle of 600 to the
horizontal. Find the ratio of the horizontal range and maximum height of A and B?

Answer 1

A body of mass $m$ is thrown with velocity $u$ at angle of $30{}^{\circ}$ to the horizontal and another body $B$ of the same mass is thrown with velocity $u$ at an angle of $60{}^{\circ}$ to the horizontal. Find the ratio of the horizontal range and maximum height of $A$ and $B$ ?

Solution.

$$m, u, \alpha_ {A} = 3 0 {}^ {\circ}, \alpha_ {B} = 6 0 {}^ {\circ};$$ $$\frac {L _ {A}}{h _ {m a x A}} -? \frac {L _ {B}}{h _ {m a x B}} -?$$

The body is thrown with velocity $u$ at angle of $\alpha$ .

$$u _ {x} = u \cos \alpha ;$$ $$u _ {y} = u \sin \alpha .$$

The max height attained.

$$h _ {m a x} = \frac {v _ {y} ^ {2} - u _ {y} ^ {2}}{- 2 g}.$$

At the max height $v_{y} = 0$ :

$$h _ {m a x} = \frac {- u _ {y} ^ {2}}{- 2 g};$$ $$h _ {m a x} = \frac {u _ {y} ^ {2}}{2 g};$$ $$h _ {m a x} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}.$$

The time of flight.

$$h = u _ {y} t - \frac {g t ^ {2}}{2};$$ $$h = u \sin \alpha t - \frac {g t ^ {2}}{2}.$$

At the end of the flight $h = 0$ :

$$0 = u \sin \alpha t - \frac {g t ^ {2}}{2};$$ $$u \sin \alpha t = \frac {g t ^ {2}}{2};$$ $$u \sin \alpha = \frac {g t}{2};$$ $$t = \frac {2 u \sin \alpha}{g}.$$

The max horizontal range of the body.

$$L = u _ {x} t;$$ $$L = u \cos \alpha t;$$ $$L = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g}.$$

A) The body A of mass $m$ is thrown with velocity $u$ at angle of $\alpha_{A} = 30{}^{\circ}$ to the horizontal.

The max height attained by the body A:

$$h _ {m a x A} = \frac {u ^ {2} \sin^ {2} \alpha_ {A}}{2 g}.$$

The max range of the body A:

$$L _ {A} = \frac {2 u ^ {2} \sin \alpha_ {A} \cos \alpha_ {A}}{g}.$$

The ratio of the horizontal range and maximum height of body A:

$$\frac {L _ {A}}{h _ {m a x A}} = \frac {2 u ^ {2} \sin \alpha_ {A} \cos \alpha_ {A} 2 g}{g u ^ {2} \sin^ {2} \alpha_ {A}} = \frac {4 \cos \alpha_ {A}}{\sin \alpha_ {A}} = \frac {4}{\tan \alpha_ {A}} = \frac {4}{\tan 30 {}^ {\circ}} = 6.9.$$

B) The body B of mass $m$ is thrown with velocity $u$ at angle of $\alpha_{B} = 60{}^{\circ}$ to the horizontal.

The max height attained by the body B:

$$h _ {m a x B} = \frac {u ^ {2} \sin^ {2} \alpha_ {B}}{2 g}.$$

The max range of the body B:

$$L _ {B} = \frac {2 u ^ {2} \sin \alpha_ {B} \cos \alpha_ {B}}{g}.$$

The ratio of the horizontal range and maximum height of body B:

$$\frac {L _ {B}}{h _ {m a x B}} = \frac {2 u ^ {2} \sin \alpha_ {B} \cos \alpha_ {B} 2 g}{g u ^ {2} \sin^ {2} \alpha_ {B}} = \frac {4 \cos \alpha_ {B}}{\sin \alpha_ {B}} = \frac {4}{\tan \alpha_ {B}} = \frac {4}{\tan 60 {}^ {\circ}} = 2.3.$$

Answer:

The ratio of the horizontal range and maximum height of body A is

$$\frac{L_A}{h_{maxA}} = 6.9$$

.

The ratio of the horizontal range and maximum height of body A is

$$\frac{L_B}{h_{maxB}} = 2.3$$

$.