Question

A girl throws a ball at an inclined wall from a height of 3 ft, hitting the wall at A with a horizontal velocity v0 of magnitude 45 ft/s. Knowing that the coefficient of restitution between the ball and the wall is 0.9 and neglecting friction, determine the distance d from the foot of the wall to the Point B where the ball will hit the ground after bouncing off the wall.

Answer 1

The diagram that describes the question is attached as a file below

Answer:

$d = 62.9 s$

Explanation:

$v_0 = 45 ft/s\\e = 0.9$

Tangent velocity, $(v_A)_t = v_0 sin (90 - 60)$

$(v_A)_t = 45 sin(30) = 22.5 ft/s$

$v_0 sin(30 - \alpha) = 22.5$....................(1)

Along the normal:

$e = \frac{0 - (v_a)_n}{(v_a)_n - 0} \\0.9 = \frac{0 - v_0 cos(30 - \alpha)}{-45 cos 30}$

$v_0 cos(30 - \alpha) = 35.07$.................(2)

Dividing equation (1) by equation (2)

$tan(30 - \alpha) = 22.5/35.07\\30 - \alpha = tan^{-1}(22.5/35.07)\\30 - \alpha = 32.68\\\alpha = - 2.68^0\\tan(\alpha - 30) = 22.5/35.07\\\alpha = 62.68^0$

$v_0 = \frac{35.07}{ cos32.69} \\v_0 = 41.67 ft/s\\h = \frac{(v_0)_y^2}{2g} \\h = \frac{(41.67sin62.68)^2}{2*32.2}\\h = 21.28 ft$

Time taken from A to E, t₁

$t_1 = \frac{(v_0)_y}{g} \\t_1 = \frac{41.67sin62.68}{32.2}\\t_1 = 1.15 s$

Time taken from E to B, t₂

$t_2 = \sqrt{\frac{2(21.28+3)}{9.8}} \\t_2 = 2.23 s$

Total time, t = 1.15 + 2.23

t = 3.38 s

$d = (v_0)_x t - CD\\d = [(41.67 cos 62.68)*3.38] - \frac{3}{tan 60} \\d = 62.9 s$

Answer 2

The distance ( d ) from the foot of the wall to the Point B where the ball hits the ground after bouncing off the wall  : 62.9

Given data :

Height = 3 ft

Horizontal velocity ( V₀ ) = 45 ft/s

Coefficient of restitution = 0.9

Determine the distance from foot of wall

Tangent velocity ( Vₐ ) t = 45 sin(30)  = 22.5 ft/s

∴ V₀sin ( 30 - ∝ ) = 22.5  ------ ( 1 )

along the normal plane

  V₀cos ( 30 - ∝ ) = 35.07  ------ ( 2 )

• step 1 : Divide equation ( 1 ) by  ( 2 )

  Tan ( 30 - ∝ ) = ( 22.5 / 35.07 )

  ∴  ∝ = 30 - 32.68

          = -2.68° ≈ 62.68°

V₀ is also expressed as :  ( 35.07 ) / ( cos 32.69 )

∴ V₀ = 41.67 ft/s

• step 2 : Determine the value of h

    h = $\frac{(Vo)^{2}y }{2*g}$  =  ( 41.67 * sin62.68 )² / ( 2 * 32.2 )

                     = 21.28 ft

• step 3 : Determine time taken from one point to another

 Time taken from point A to E

T₁ = ( 41.67 * sin62.68 ) /  32.2

    = 1.15 secs

 Time from point E to B

T₂ = $\sqrt{\frac{2(21.28+3)}{9.8} }$  =  2.23 secs

 

Therefore the Total time taken = 2.23 + 1.15 = 3.38 secs

Final step : Determine the distance d

d = ( V₀ )x t - CD

  = [ ( 41.67*cos62.68 ) * 3.38 ] -  $\frac{3}{tan 60}$

  = 62.9

Hence we can conclude that The distance ( d ) from the foot of the wall to the Point B where the ball hits the ground after bouncing off the wall is 62.9

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