Question

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​

Answer 1

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

Answer 2

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