Question

Question : A cannon fires successively two shells with velocity $\sf V_{o} = 250 m/s$ ; The first at the angle $\sf \theta_{1} = 53°$ and the second at an angle of $\sf \theta_{2} = 37°$ to the horizontal, in the same vertical plane, neglecting the air drag, find the time interval (in seconds) between firings leading to the collision of the shells (take g = 10m/s²)

Answer : 10 sec ​

Answer 1

Explanation:

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Answer 2

$\large \dag$ Question :-

A cannon fires successively two shells with velocity $\sf V_{o} = 250 m/s$ ; The first at the angle $\sf \theta_{1} = 53°$ and the second at an angle of $\sf \theta_{2} = 37°$ to the horizontal, in the same vertical plane, neglecting the air drag, find the time interval (in seconds) between firings leading to the collision of the shells (take g = 10m/s²)

$\large \dag$ Answer :-

Time interval between firings is 10 seconds

$\large \dag$ Step by step Solution :-

Firslty Let's calculate the velocities of both particles in both x and y directions.

☆ For Particle 1 》

Let,

• velocity in x - direction be = $\sf u_{x{_{_1}}}$

• velocity in y - direction be = $\sf u_{y{_{_1}}}$

Now,

$\sf u_{x{_{_1}}} = \sf V_{o}.cos53 \degree$

$:\longmapsto \rm u_{x{_{_1}}} = 250 \times \frac{3}{5}$

$:\longmapsto \rm \underline{ \underline{u_{x{_{_1}}} = 150 \: m/s}}$

Also,

$\sf u_{y{_{_1}}} = \sf V_{o}.sin53 \degree$

$:\longmapsto \rm u_{x{_{_1}}} = 250 \times \frac{4}{5}$

$:\longmapsto \rm \underline{ \underline{u_{x{_{_1}}} = 200 \: m/s}}$

☆ For Particle 2 》

Let,

• velocity in x - direction be = $\sf u_{x{_{_2}}}$

• velocity in y - direction be = $\sf u_{y{_{_2}}}$

Now,

$\sf u_{x{_{_2}}} = \sf V_{o}.cos37\degree$

$:\longmapsto \rm u_{x{_{_2}}} = 250 \times \frac{4}{5}$

$:\longmapsto \rm \underline{ \underline{u_{x{_{_2}}} = 200 \: m/s}}$

Also,

$\sf u_{y{_{_2}}} = \sf V_{o}.sin37 \degree$

$:\longmapsto \rm u_{y{_{_2}}} = 250 \times \frac{3}{5}$

$:\longmapsto \rm \underline{ \underline{u_{y{_{_2}}} = 150 \: m/s}}$

Now Let's calculate time taken by both particles to reach point P.

Let,

• Time taken by 1st particle be = $\sf t _{1}$

• Time taken by 2nd particle be = $\sf t _{2}$

❒ Taking x - direction of Both Particles :-

In X direction acceleration of both particles will be zero.

☆ For Particle 1 》

Applying 2nd Eq. of kinematics ;

$:\longmapsto \sf S_{x{_{_1}}} =u _{x{_{_1}}}t _{1} + \frac{1}{2} \times a_{x{_{_1}}} {t _{1}}^{2}$

$:\longmapsto \sf S_{x{_{_1}}} = u_{x{_{_1}}}t _{1}+0$

$:\longmapsto \sf x = 150 \times t _{1}$

$:\longmapsto \sf\underline{ \underline{ t _{1} = \frac{x}{150}}}\:\: ----(1)$

☆ For Particle 2 》

Applying 2nd Eq. of kinematics ;

$:\longmapsto \sf S_{x{_{_2}}} =u _{x{_{_2}}}t _{2} + \frac{1}{2} \times a_{x{_{_2}}} {t _{2}}^{2}$

$:\longmapsto \sf S_{x{_{_2}}} = u_{x{_{_2}}}t _{2}+0$

$:\longmapsto \sf x = 200 \times t _{2}$

$:\longmapsto \sf\underline{ \underline{ t _{2} = \frac{x}{200}}} \:\: ----(2)$

❒ Taking y - direction of Both Particles :-

in Y direction acceleration of both particles is - g.

☆ For Particle 1 》

Applying 2nd Eq. of kinematics ;

$:\longmapsto \sf S_{y{_{_1}}} = u_{y{_{_1}}}t _{1} - \frac{1}{2} g {t _{1}}^{2}$

⏩ Putting values ;

$:\longmapsto \sf y = 200 \times \frac{x}{150} - \frac{1}{2} \times 10 \times \bigg( \frac{x}{150} \bigg) ^{2}$

$:\longmapsto \sf y = \frac{4}{3} x - \frac{1}{2} \times 10 \times \frac{ {x}^{2} }{22500}$

$:\longmapsto \sf y = \frac{4}{3} x - \frac{ {x}^{2} }{4500} \: \: - - - - (3)$

☆ For Particle 2 》

Applying 2nd Eq. of kinematics ;

$:\longmapsto \sf S_{y{_{_2}}} = u_{y{_{_2}}}t _{2} - \frac{1}{2} g {t _{2}}^{2}$

⏩ Putting values ;

$:\longmapsto \sf y = 150 \times \frac{x}{200} - \frac{1}{2} \times 10 \times \bigg( \frac{x}{200} \bigg) ^{2}$

$:\longmapsto \sf y = \frac{3}{4} x - \frac{1}{2} \times 10 \times \frac{ {x}^{2} }{40000}$

$:\longmapsto \sf y = \frac{3}{4} x - \frac{ {x}^{2} }{8000} \: \: - - - - (4)$

⏩ Equating value of y from (3) and (4) ;

$:\longmapsto\sf \frac{4}{3} x - \frac{ {x}^{2} }{4500} = \frac{3}{4}x - \frac{ {x}^{2} }{8000}$

$:\longmapsto \sf \frac{x}{4500} - \frac{x}{8000} = \frac{4}{3} - \frac{3}{4}$

⏩ On Simplifying further ;

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf x = 6000} }}}$

⏩ Putting Value of x in (1) and (2) we get ;

$:\longmapsto \sf t _{1} = \frac{6000}{150}$

$\orange{ \large :\longmapsto  \underline {\boxed{{\bf t _{1} = 40} }}}$

Also,

$:\longmapsto \sf t _{2} = \frac{6000}{200}$

$\orange{ \large :\longmapsto  \underline {\boxed{{\bf t _{2} = 30} }}}$

Therefore,

Time interval between firings leading to the collision of the shells is :

$\sf \: \: \Delta t = t_1-t_2$

$:\longmapsto \sf \Delta t = 40 - 30$

$\blue{ \large :\longmapsto  \underline {\boxed{{\bf \Delta t = 10 \: s} }}}$

Hence Time interval between firings leading to the collision of the shells is 10 second.