. A boy wishes to throw a ball through a house via two small openings, one in the front and the other in the back window, the second window being directly behind the first. If the boy stands at a distance of 5m

in front of the house and the house is 6m deep and if the opening in the front window is 5m above him

and that in the back window 2m higher, calculate the velocity and the angle of projection of the ball that

will enable him to accomplish his desire.

1) The angle of projection of the ball is approximately 52.496° above the horizontal

2) The velocity with which the boy throws the ball to enable him accomplish his desire is approximately 14.769 m/s

Explanation:

In the projectile motion of the ball, the parameters are;

The distance the boy stands from the front of the house = 5 m

The depth of the house, front to back = 6 m

The height of the opening of the front window above him = 5 m

The height of the back window = 2 m higher than the front window

Therefore, the height of the back window = 5 m + 2 m = 7 m

1) The general formula for the vertical height of a parabolic projectile motion is given as follows;

y = a·x² + b·x + c

At y = 0, x = 0, therefore, c = 0

At y = 5, x = 5, we have;

5 = a·5² + b·5

∴ 5 = 25·a + 5·b...(1)

At y = 7, x = 11, we have;

7 = a·11² + b·11

7 = 121·a + 11·b...(2)

Making 'a' the subject of equation (1) and (2) and equating both values of 'a' gives;

For equation (1);

a = (5 - 5·b)/25 = 1/5 - b/5

a = 1/5 - b/5

For equation (2);

a = (7 - 11·b)/121 = 7/121 - b/11

a = 7/121 - b/11

∴ 1/5 - b/5 = 7/121 - b/11

1/5 - 7/121 = b/5 - b/11

86/605 = 6·b/55

b = (86/605) × (55/6) = 43/33

b = 43/33

a = 1/5 - b/5 = 1/5 - (43/33)/5 = -2/33

∴ y = (-2/33)·x² + (43/33)·x

The slope of the curve = dy/dx = d((-2/33)·x² + (43/33)·x)/dx = -4/33·x + 43/33

The slope of the curve at any point on the projectile = -4/33·x + 43/33

The slope at the origin is given by plugging x = 0  as follows

The slope at the origin = -4/33 × 0 + 43/33 = 43/33

The slope at the origin = tan(θ) = 43/33

Where;

θ = The angle of projection of the ball

∴ θ = arctan(43/33) ≈ 52.496°

The angle of projection of the ball, θ ≈ 52.496° above the horizontal

2) At the back widow, the equation for the vertical height, 'h', is given as follows;

Where;

h = The vertical height of the back window = 7 m

tan∠θ = 43/33

x = The horizontal distance of the back window from the boy = 5 m + 6 m = 11 m

g = The acceleration due to gravity = 9.8 m/s²

v₁ = The velocity with which the boy throws the ball

cos²∠θ = cos²(arctan(43/33)) = 1089/2938

Plugging in the values gives;

The velocity with which the boy throws the ball to enable him accomplish his desire, v₁ ≈ 14.769 m/s.

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it has to be b, about 97 percent.