if the time of flight is 5 sec and range is 100m then find the angel of projection ,horizontal range and the vertical range
Answers
Let the angle of projection of the bullet from the horizontal be θ .
Let the initial velocity be “u”.
The vertical component of this velocity would be usinθ.
When the bullet reaches the maximum height the vertical component of its velocity is “0”.
Therefore,
0= usinθ-gt
Where g is acceleration due to the gravity and “t” is time taken to attain the maximum height.
Therefore
t= usinθ/g
The bullet will take same time “t” in its downward movement as it had taken to go up.
Therefore
T=2t = 2usinθ/g
Or
u= Tg/2sinθ
The horizontal velocity of the bullet is ucosθ.
Therefore in time “T” it would cover a distance of
Tucosθ which is same as “R”
Therefore
R= Tucosθ
Using the value of u as determined above,
R= T x (Tg/2sinθ) x cosθ
R= T^2 g /2tanθ
tanθ= T^2g/2R
θ= arctan ( T^2g/2R)
hope it was helpfull
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