[tex]\huge{\orange{\underline{\underline{Question:-}}}}[/tex]
When a particle is projected at some angle to the horizontal it has a range r and time of flight t1 if the same particle is projected with the same speed at some other angle to have the same range its time of flight is t2 then :-
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Option - C
Let the projected velocity be u and range be R and time taken by the projected body be t1 and t2 with angle .
Let the angle in first case be and second case be
Time period of projectile is :-
Verified answer
Answer:
3.) t1•t2 = 2R/g
Step by step explanations :
Method 1
Given that,
Maximum range of a projectile at two different angles at the same speed are equal.
We know that,
Maximum Range of a projectile at a vertical angle @
and Maximum range at angle (90 - @ ) are equal
given the time of flight at angle @ = t1
so,
t1 = ([email protected])/g
and time of flight at angle (90 - @) = t2
so,
t2 = 2usin(90 - @) /g
t2 = [email protected]/g
Now,
t1 • t2
([email protected]/g)([email protected]/g)
= (2u² [email protected] [email protected] )/g²
= 2u²[email protected]/g² [[email protected] [email protected] = [email protected]]
= 2R/g
[R = u² [email protected]/g]
so,
t1•t2 = 2R/g
_____________
Method 2
Refer to the attachment