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Amanda [17]
3 years ago
11

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

Physics
1 answer:
SVEN [57.7K]3 years ago
6 0

Answer:

a) Knowns, initial speed v_{i}=13.0 m/s, final speed v_{f}=0 m/s and gravity due it is a constant g=9.8m/s^{2}

b) The maximum high reached by the dolphin is y_{max}=8.62 m

c) Total time is t=2.65s

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at 0m/s speed.

b) Second, now that we know final speed we use v_{f} =v_{i}-gt, as we clear for t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s.

Then we use y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2}  =8.62m

c)Third, finally we can use y=v_{i}t-\frac{1}{2} gt^{2}, as we know y=0m when the dolphin fall into the water again and v_{i} =13.0m/s, then we have 0=(13m/s)t-\frac{1}{2} (9.8m/s^{2}  )t^{2} is a quadratic form 0=t(13.0-4.9t) so we have t_{i}=0s and t_{f}=\frac{13}{4.90}  =2.65s

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