Question

. A boy wishes to throw a ball through a house via two small openings, one in the front and the other in

Answer 1

Answer:

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;

$h = (tan \angle \theta) \cdot x - \left(\dfrac{g}{2 \cdot v_1^2 \cdot cos^2 \angle \theta } \right )\cdot x^2$

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;

$7 = \left (\dfrac{43}{33} \right) \times 11 - \left(\dfrac{9.8}{2 \cdot v_1^2 \cdot \dfrac{1089}{2938} } \right )\cdot 11^2$

$\therefore v_1^2 = \dfrac{71981}{330}$

$\therefore v_1 = \sqrt{\dfrac{71981}{330}} \approx \pm14.769$

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