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just olya [345]
3 years ago
10

A dolphin leaps out of the water at an angle of 36.6° above the horizontal. The horizontal component of the dolphin's velocity i

s 7.87 m/s. Calculate the magnitude of the vertical component of the velocity.
Physics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

Explanation:

Given

Dolphin leaps out an angle \theta =36.6^{\circ}

Horizontal component of dolphin velocity u_x=7.87\ m/s

Suppose u is the launch velocity of dolphin

therefore u\cos \theta =u_x---1

and vertical velocity u_y=u\sin \theta ----2

divide 1 and 2 we get

\tan \theta =\frac{u_y}{u_x}

u_y=u_x\tan \theta

u_y=7.87\cdot \tan (36.6)

u_y=5.84\ m/s

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Compare the pressure exerted by the liquid at points A, B and C. Justify your answer
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2 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
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Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

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