Physics, asked by emmanuelmeta851, 1 year ago

A boy wishes to throw a ball through a house via two small openings one in the front and the other in back window the second window being directly behind the first if the boy stand at a distance of 5m in front of the house and the house is 6m deep and if the opeing in front window is 5m above him and that in the back window 2m higher Calculate the velocity and angle of projection of the ball that will enable him to accomplish his desire.

Answers

Answered by amitnrw

Answer:

Explanation:

If We Understand this Question

The Horizontal Distance = 5 m When Vertical Distance = 5 m

& Horizontal Distance = 5 + 6 = 11m When Vertical Distance = 5+2 = 7 m

Now Let say Velocity = V & angle = α

Horizontal Speed = VCosα

Vertical Speed = VSinα

Let say at T₁ time Ball crosses front window

then Horizontal Distance = VCosα*T₁ = 5 m => T₁ = 5/ VCosα

& Vertical Distance = VSinαT₁ - (1/2)gT₁² = 5

Using g = 10 & T₁ = 5/ VCosα

=> VSinα(5/ VCosα) - 5 (5/ VCosα)² = 5

=> Tanα - 25/V²Cos²α = 5

=> Tanα = 5 + 25/V²Cos²α

Let say at T₂ time Ball crosses back window

then Horizontal Distance = VCosα*T₂ = 11 m => T₂ = 11/ VCosα

& Vertical Distance = VSinαT₂ - (1/2)gT₂² = 7

Using g = 10 & T₂ = 11/ VCosα

=> VSinα(11/ VCosα) - 5 (11/ VCosα)² = 7

=> 11Tanα - 605/V²Cos²α = 7

=> Tanα = (7 + 605/V²Cos²α)/11

5 + 25/V²Cos²α = (7 + 605/V²Cos²α)/11

=> 55 + 275/V²Cos²α = 7 + 605/V²Cos²α

=> 48 = 330/V²Cos²α

=> V²Cos²α = 55/8

putting V²Cos²α = 55/8 in below

Tanα = 5 + 25/V²Cos²α

=> Tanα = 5 + 25/(55/8)

=>Tanα = 5 + 40/11

=> Tanα = 95/11

=> α = Tan⁻¹(95/11) = 83.4°

Cosα = 11 / √(11² + 95²)

=> Cos²α = 121/9146

V²Cos²α = 55/8

=> V² (121/9146) = 55/8

=> V² = 22865/44

=> V = 22.8 m/s

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