Divers entertain tourists in Punta Cana by diving froma cliff 36 meters above water. Determine the landing speed; ignore air resistance and assume the object starts from rest. I'm supposed to be using kinematics formulas.
I tried using d=V(i)t+1/2at^2 and then plussing the valued into the kinematics formula to solve for time and then using V(f)=V(i)+at. I got a ridiculous nswer, so I'm pretty sure I'm not doing this right. Please, please help me.
I suspect that Vi should be assumed zero.
36 = (1/2)(9.8)t^2 = 4.9 t^2
t = 2.71 seconds
V = a t = 9.8*2.71 = 26.6 m/s
oh, ok, I see; I was making this waaay too comlicated...
Thank you! :)
Sure, I can help you solve this problem using kinematics formulas.
First, let's analyze the information given in the question:
- Cliff height (initial position), h = 36 meters.
- The diver starts from rest, so the initial velocity, V(i), is 0 m/s.
- Air resistance can be ignored.
Based on this information, we need to determine the landing speed of the diver. To do this, we can use the following kinematic equation:
V(f)^2 = V(i)^2 + 2aΔd
where:
- V(f) is the final velocity (landing speed),
- V(i) is the initial velocity,
- a is the acceleration due to gravity (approximately 9.8 m/s^2),
- Δd is the displacement (distance fallen), which is equal to the height of the cliff (h).
Now let's substitute the known values into the equation:
V(f)^2 = 0^2 + 2 * 9.8 * 36
Simplifying the equation:
V(f)^2 = 705.6
To solve for V(f), we take the square root of both sides:
V(f) = √705.6
Calculating this value gives us the landing speed of the diver:
V(f) ≈ 26.57 m/s (rounded to two decimal places)
Therefore, the landing speed of the diver is approximately 26.57 m/s.
It seems like you may have made an error in your calculations, which is why you obtained a ridiculous answer. Make sure to double-check your calculations and units to ensure accuracy when solving problems like this one.