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2. Samuel leaves the end of a 4.0-m-high diving board and strikes the
water 1.3 s later, 3.0 m beyond the end of the board. Considering the
diver as a particle, determine:
(a) His initial velocity,
(b) The maximum height reached;
(c) The velocity Vf
with which he enters the water.

Respuesta :

Answer:

a) v = 4.9 m/s at 62° above the horizontal

b) h = 0.94 m above the board or 4.94 m above the water

c) v = 10 m/s at 77° below the horizontal

Explanation:

Assume UP and FOREWARD are positive and the diving board is the origin.

Ignore air resistance.

s = s₀ + v₀t + ½at²

In the vertical

-4 = 0 + vy₀(1.3) + ½(-9.8)(1.3²)

vy₀ = 4.281 m/s

in the horizontal

3.0 = 0 + vx₀(1.3) + ½(0)(1.3²)

vx₀ = 2.307... m/s

v = √(4.281² + 2.307²) = 4.863...

tanθ = 4.281 / 2.307

θ = 61.680...

v² = u² + 2as

s = (v² - u²) / 2a

h = (0² - 4.281²) / 2(-9.8) = 0.9350...

v² = 4.863² + 2(9.8)(4)

v = 10.1019...

cosθ = 2.307/10.1019

θ = 76.798

onyebuchinnaji

(a) The initial velocity of Samuel is 4.02 m/s

(b) The maximum height reached by Samuel is 0.55 m

(c)  The final velocity when he enters the water is 5.20 m/s

The given parameter:

  • the height of the board, h = 4 m
  • time to strike the water, t = 1.3 s
  • the horizontal distance traveled, x = 3.0 m

(a) The initial velocity of Samuel is calculated as;

[tex]h = v_o_y t + \frac{1}{2} gt^2\\\\[/tex]

Assume downward motion to be in negative direction

The vertical component of the velocity is calculated as;

[tex]4 = 1.3\times v_0_y + 0.5\times 9.8\times 1.3^2\\\\4 = 1.3v_{0y} + 8.281\\\\1.3v_{0y} = 4- 8.281\\\\1.3v_{0y} = -4.281\\\\v_{0y} = \frac{-4.281}{1.3} \\\\v_{0y} = -3.293 \ m/s\\\\v_{0y} = 3.293 \ m/s \ \ \ (downwards)[/tex]

The horizontal component of the velocity is calculated as;

Assume forward motion to be in positive direction

[tex]x = v_{0x}t \\\\3 = 1.3v_{0x}\\\\v_{0x} = \frac{3}{1.3} \\\\v_{0x} = 2.308 \ m/s[/tex]

The resultant of initial velocity is calculated as;

[tex]v_0 = \sqrt{v_{0y}^2 + v_{0x}^2} \\\\v_0 = \sqrt{3.293^2 + 2.308^2} \\\\v_0 = \sqrt{16.17} \\\\v_0 = 4.02 \ m/s[/tex]

(b) The maximum height reached by Samuel is calculated as;

[tex]v_f_y^2 = v_0_y^2 + 2gh\\\\at \ maximum \ height \ v_{fy} = 0\\\\0 = v_0_y^2 + 2gh\\\\-2gh = v_0_y^2\\\\h = \frac{v_0_y^2}{-2g} \\\\h = \frac{(3.293)^2}{-2\times 9.8} \\\\h = -0.55 \ m\\\\h = 0.55 \ m \ (downwards)[/tex]

(c) The final velocity when he enters the water;

The height traveled before he hits the water, h = 0.55 m

[tex]v_f^2 = v_{0}^2 + 2gh\\\\v_f^2 = (4.02)^2 + (2\times 9.8\times 0.55)\\\\v_f^2 = 26.94\\\\v_f = \sqrt{26.94} \\\\v_f = 5.20 \ m/s[/tex]

Learn more here: https://brainly.com/question/13372763

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