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A diver 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 the maximum height reached (I solved for the initial velocity and degree above the horizontal. What do I do next? I have been looking at the big 5 for ever and I’m still stuck.)

Respuesta :

let the diver has vertical speed as vy

now we can use kinematics to find out the vertical speed

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

as we know that

[tex]-4 = v_y (1.3) - \frac{1}{2}(9.8)(1.3^2)[/tex]

by solving above equation we have

[tex]v_y = 3.3 m/s[/tex]

now for finding the maximum height we will use kinematics equation

[tex]v_f^2 - v_y^2 = 2 a y[/tex]

[tex]0 - 3.3^2 = 2(-9.8)y[/tex]

[tex]y = 0.55 m[/tex]

so the diver will move to 0.55 m high from his initial position

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