a projectile is fired with velcotiy u
Answers
★Given :-
- A projectile is fired with a velocity u making an angle theta with the horizontal.
★To find :-
- Expression for time of flight.
- Condition for maximum horizontal range.
★Solution :-
(i) Time of flight :
✦Total time interval which involves the time of ascent and time of descent of a projectile is known as time of flight.
✦It depends on the initial velocity of the object and the angle of the projection, θ .
The initial velocity of the projectile is u and angle with the x - axis is θ.
- uₓ is u cosθ.
- uᵧ is u sinθ
As the projectile starts from the ground and returns to the ground,the vertical displacement is zero.
From the second law of motion :
✦S = ut + 1/2at²
Putting values,
→S = ut + 1/2at²
→0 = ut + 1/2 (-g)t²
→ut = 1/2 gt²
→u = 1/2 gt
→2uᵧ /g = t
→ t = 2usinθ/g
Therefore,
Time of flight = 2u sinθ/g
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(ii)Condition for maximum horizontal range :
Horizontal displacement of a projectile along x-axis is called as horizontal range.
✦R = u²sin2θ/g
Range will be maximum when,
→sin2θ = 1
→sin 90°
Therefore,
→ 2θ = 90°
→ θ = 90/2
→ θ = 45°
Hence,
Horizontal range is maximum when it is projected at an angle of 45°.
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