Question

A projectile is launched from a point on horizontal plane

Answer 1

Answer:

√2 Seconds

Take it's velocity components

in x direction

$u \cos( \alpha ) = 10 \sqrt{2} \times \frac{1}{2} = 5 \sqrt{2}$

Which will be constant .

Now in y direction

$u \sin( \alpha ) = 10 \sqrt{2} \times \frac{ \sqrt{3} }{2} = 5 \sqrt{6}$

It will change over time .

After time t , velocity of particle will be

$5 \sqrt{6} - gt = 5( \sqrt{6} - 2t)$

And height

$s =h = 5 \sqrt{6} t - \frac{1}{2} g {t}^{2} \\ \\ h = 5 (\sqrt{6} t - {t}^{2} )$

Total kinetic energy in system is equal to potential energy

$\frac{1}{2} m( {v_x}^{2} + {v_y}^{2}) = mgh \\ \\ ( 5\sqrt{2} ) {}^{2} + 5 {}^{2} ( \sqrt{6} - 2t) {}^{2} = 2 \times 10 \times 5( \sqrt{6} t - {t}^{2} ) \\ \\ 2 + 6 + 4 {t}^{2} - 4 \sqrt{6} t = 4 \sqrt{6}t - 4 {t}^{2} \\ \\ 8 {t}^{2} - 8 \sqrt{6}t + 8 = 0 \\ \\ {t}^{2} - \sqrt{6} t + 1 = 0$

It is quadratic , it ' s two value will give us two times when both energies are equal .

But we need difference between them,

or we need difference of roots

$| \alpha - \beta | = \sqrt{( \alpha + \beta ) {}^{2} - 4 \alpha \beta } \\ | \alpha - \beta | = \sqrt{6 - 4} = \sqrt{2}$

Answer 2

Answer:

this is your answer

Explanation:

please mark brilliant