To solve this problem, we’ll divide it into two parts: first determining the time the divers are in the air, and then using that time to calculate the necessary horizontal velocity.
Part 1: Time in Air
When the divers jump from the cliff, they start with an initial vertical velocity of 0 m/s. The initial position in the vertical direction is also 0. We need to find the time it takes for them to fall the 41 meters height, under the acceleration due to gravity.
We can use the following kinematic equation for uniformly accelerated motion to find the time:
[tex]\[ d = V_{i}t + \frac{1}{2}at^2 \][/tex]
Here, [tex]\( d \)[/tex] is the distance fallen (41 m), [tex]\( V_{i} \)[/tex] is the initial vertical velocity (0 m/s), [tex]\( a \)[/tex] is the acceleration due to gravity (9.81 m/s^2), and [tex]\( t \)[/tex] is the time in seconds.
Plugging in our known values:
[tex]\[ 41 = 0 \cdot t + \frac{1}{2} \cdot 9.81 \cdot t^2 \][/tex]
[tex]\[ 41 = 4.905t^2 \][/tex]
Now we solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{41}{4.905} \][/tex]
[tex]\[ t^2 = 8.36 \][/tex]
[tex]\[ t = \sqrt{8.36} \][/tex]
[tex]\[ t = 2.89 \, \text{seconds} \][/tex]
Part 2: Minimum Horizontal Velocity
Now that we know the divers are in the air for approximately 2.89 seconds, we can use this time to calculate the minimum horizontal velocity needed to clear the 20 meters of rocks.
Horizontal velocity is calculated using the formula:
[tex]\[ v = \frac{d}{t} \][/tex]
Here, [tex]\( d \)[/tex] is the distance to clear (20 m), and [tex]\( t \)[/tex] is the time in the air (2.89 s).
[tex]\[ v = \frac{20}{2.89} \][/tex]
[tex]\[ v = 6.92 \, \text{m/s} \][/tex]
To get an answer suitable for the divers (as they can't achieve a precise non-integer velocity), we'll round this to the nearest whole number:
[tex]\[ v \approx 7 \, \text{m/s} \][/tex]
So, the minimum horizontal velocity the La Quebrada cliff divers must jump with to clear the 20 meters of rocks is approximately 7 m/s when rounded to the nearest whole number.
