What is the Minimum Horizontal Velocity for a Safe Dive from a 61m High Cliff?

[Larrytsai]

Please just check my answer.

Divers at Acapulco dive from a cliff that is 61m high. If the rocks below the cliff extend outward for 23m, what is the minimum horizontal velocity a diver must have to clear the rocks safely?

x problem

x=Vx(t)
23=Vx(t)
23=Vx(3.52)
23/3.52=Vx
Answer is Vx=6.52m/s

Y problem

Vy=0m/s
t=?
d=-61m
a=-9.8m/s/s

d=1/2a(txt)+Vi(t)

-61=-4.9(txt)
-61/-4.9=txt
12.44=txt
square root 12.44=txt Answer for Time is 3.52 Seconds

 

[Shooting Star]

All correct.

 

[ogb p]

I would first like to commend you for your efforts in solving this problem. However, I would like to point out a few things in your solution. Firstly, it is important to mention the units in your final answer. In this case, the horizontal velocity should be expressed as 6.52 m/s. Additionally, it is important to clarify what the variables Vx and Vy represent. In this case, Vx represents the horizontal velocity and Vy represents the vertical velocity.

Furthermore, in your Y problem, it is not necessary to find the time (t) since it is already given in the X problem. However, if you were to find the time, it would be important to include the units as well.

Lastly, I would like to mention that it is always a good idea to double-check your calculations and make sure they are accurate. Overall, your solution is correct and well thought out. Keep up the good work!