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2 years ago

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A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 13.2 m/s

and her body makes an angle of 81.8 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

1 answer:

kodGreya [7K]2 years ago

5 0

Answer:

a) u = 9.88 m/s

b) $\theta = 79^0$

Explanation:

given,

speed of the diver = 13.2 m/s

Angle made w.r.t horizontal = 81.8 °

The horizontal component of final velocity

v_h = 13.2 cos 81.8 °

The vertical component of final velocity

v_v = 13.2 sin 81.8 °

using equation of motion

initial velocity of vertical component

$u_v = \sqrt{v_v^2-2as}$

$u_v = \sqrt{(13.2 sin 81.8^0)^2-2(-9.8)(-3.9)}$

= 9.70 m/s

the horizontal component of the initial velocity

u_h = v_h = 13.2 cos 81.8 ° = 1.88 m/s

magnitude

$u = \sqrt{9.7^2+1.88^2}$

u = 9.88 m/s

direction

$\theta = tan{-1}(\dfrac{9.7}{1.88})$

$\theta = 79^0$

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