Find the range of ball which, when projected with a velocity of 29.4 m s ^- 1 just passes over a pole 4.9 m high
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Find the range of ball which, when projected with a velocity of 29.4 m s ^- 1 just passes over a pole 4.9 m high
plz friends give me answer plz
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Verified answer
Answer :-
Range of the ball is 55 metres .
Explanation :-
We have :-
→ Velocity of projection = 29.4 m/s
→ Height of pole = 4.9 m
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From the formula of maximum height of a projectile, we have :-
h = u²sin²θ/2g
⇒ 4.9 = [(29.4)² × sin²θ]/2(9.8)
⇒ 4.9 = [864.36 × sin²θ]/19.6
⇒ 864.36 × sin²θ = 19.6(4.9)
⇒ sin²θ = 96.04/864.36
⇒ sin²θ = 1/9
⇒ sin θ = 1/3
⇒ sin θ = 0.33
We have sin θ = 1/3 , now solving for cos θ, we get :-
⇒ cos θ = √8/3
⇒ cos θ = 0.943
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Now from the formula of range of a projectile :-
R = u²sin2θ/g
⇒ R = u²(2sinθ)(cosθ)/g
⇒ R = (29.4)²(2 × 0.33)(0.943)/9.8
⇒ R = (864.36)(0.66)(0.943)/9.8
⇒ R = 537.96/9.8
⇒ R = 54.9 ≈ 55 m
Given :-
Velocity = 29.4 m/s
Height = 4.9
Acceleration = 10 m/s²
To Find :-
Range
Solution :-
We know that
Height = (u)² × sin²θ/2 × g
4.9 = (29.4)² × sin²θ/2 × 10
4.9 = 864.36 × sin²θ/20
4.9 × 40 = 864.36 × sin²θ
49/10 × 40 = 864.36 × sin²θ
49 × 4 = 864.36 × sin²θ
196 = 864.36 × sin²θ
196/864.36 = sin²θ
1/9 = sin²θ
√(1/9) = sinθ
1/3 = sinθ
Now,
√8/3 = cosθ
Range = u² × sin2θ/g
R = u² × sinθ × 2 × cosθ/g
R = (29.4)² × (1/3 × 2) × (√8/3)/10
R = 864.36 × (2/3 × √8/3)/10
R = 86436 × (2√8/3)/1000
R = 86.436 × 2√8/3
R = 28.812 × 2√8
R = 55 m