Question 5: Motion in 2D & Vectors

Question

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are 5 s and 10 s , respectively, then the value of $R$ is
$\_\_\_\_$ m.  (Take $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

For two projectiles launched with the same initial speed u to have the same horizontal range R, their angles of projection must be complementary.
Let the first angle be $	heta$. Then the second angle must be ( $90^{\circ}-	heta$ ) .
The time taken for the projectile to return to the same horizontal level.
$$\mathrm{T}=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{~g}} \ldots (i) $$
And the total horizontal distance covered is the range of projectile, R .
$$ R=\frac{u^2 \sin (2 	heta)}{g}=\frac{2 u^2 \sin 	heta \cos 	heta}{g} $$
$$\Rightarrow $$ $$\mathrm{R}=\frac{2 \mathrm{u}^2 \sin 	heta \cos 	heta}{\mathrm{~g}}........ (ii) $$
For the first body $\left(\mathrm{T}_1=5 \mathrm{~s}ight)$ :
$$ \mathrm{T}_1=\frac{2 \mathrm{u} \sin 	heta}{\mathrm{~g}} \Rightarrow \sin 	heta=\frac{\mathrm{g} \mathrm{~T}_1}{2 \mathrm{u}} $$
For the second body ( $\mathrm{T}_2=10 \mathrm{~s}$ ):
$$ \mathrm{T}_2=\frac{2 \mathrm{u} \sin \left(90^{\circ}-	hetaight)}{\mathrm{g}} \Rightarrow \cos 	heta=\frac{\mathrm{g} \mathrm{~T}_2}{2 \mathrm{u}} $$
$\Rightarrow \mathrm{T}_1 	imes \mathrm{T}_2=\left(\frac{2 \mathrm{u} \sin 	heta}{\mathrm{g}}ight) 	imes\left(\frac{2 \mathrm{u} \cos 	heta}{\mathrm{g}}ight)$
$$\Rightarrow $$ $\mathrm{T}_1 \mathrm{~T}_2=\frac{2}{\mathrm{~g}}\left(\frac{2 \mathrm{u}^2 \sin 	heta \cos 	heta}{\mathrm{~g}}ight)$
Using the relation (ii)
$$ \mathrm{T}_1 \mathrm{~T}_2=\frac{2 \mathrm{R}}{\mathrm{~g}} $$
$$\Rightarrow $$ $$ \mathrm{R}=\frac{\mathrm{g} \mathrm{~T}_1 \mathrm{~T}_2}{2} $$
Here, the time of flight of first projectile is $\mathrm{T}_1=5 \mathrm{~s}$ and that of second projectile is it $\mathrm{T}_2=10 \mathrm{~s}$ Putting the values in the formula of horizontal range,
$$ \mathrm{R}=\frac{10 	imes 5 	imes 10}{2} $$
$$\Rightarrow $$ $$ \mathrm{R}=250 \mathrm{~m} $$
Therefore, the value of the horizontal range R is 250 m.
Hence, the correct option is (A).

Answer

250

Answer

25

Answer

500

Answer

125

Motion in 2D & Vectors

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range R. If the time of flight of these bodies are 5 s and 10 s , respectively, then the value of R is \_\_\_\_ m. (Take g=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: