Question

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.
(a) List the knowns in this problem.
(b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.
(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.

Answer 1

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 $t_{f}=\frac{13}{4.90}  =2.65s$