Question

There are two values of time for which a projectile is at the same height. The sum of these two times is equal to (T= time of flight of the projectile)

Answer 1

Given:

There are two values of time for which a projectile is at the same height.

To find:

Sum of the two times.

Calculation:

Let that specified height be h ;

$h = u_{y}t - \dfrac{1}{2} g {t}^{2}$

$= > h = \{u \sin( \theta) \} t - \dfrac{1}{2} g {t}^{2}$

$= > 2h = 2\{u \sin( \theta) \} t - g {t}^{2}$

$= > g {t}^{2} - 2u \sin( \theta) t + 2h = 0$

Let the two roots be t1 and T2 when this height h is obtained ; Therefore sum of roots will be :

$\therefore \: t1 + t2 = - \dfrac{b}{a}$

$= > \: t1 + t2 = - \bigg( \dfrac{ - 2u \sin( \theta) }{g} \bigg)$

$= > \: t1 + t2 = \dfrac{ 2u \sin( \theta) }{g}$

$\boxed{ \sf{ = > \: t1 + t2 = \dfrac{ 2u \sin( \theta) }{g} = T}}$

Hope It Helps.

Answer 2

Answer: T

Explanation:

Time taken to go from O to A = t1

Time taken to go from O to B = t2

Also ,

Time taken to go from O to A = time taken to go from B to D

Therefore, t1 +t2 = t2 + time taken to go from B to D

= T

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