Divers dive straight out from a cliff that is 21 m high. If the rocks below the cliff extend outward for 15 m, what is the minimum horizontal velocity a diver must have to clear the rocks safely
h = Vo*t + 0.5g*t^2 = 21 m.
0 + 4.9t^2 = 21,
t^2 = 4.29,
t = Tf = 2.07s. = Fall time = Time in flight.
d = Vt,
V = d/t = 15 / 2.07 = 7.25m/s. = Hor.
velocity.
To determine the minimum horizontal velocity required for the diver to clear the rocks safely, we can use the principles of projectile motion. Let's break down the problem into two components: vertical and horizontal.
First, let's consider the vertical motion. The initial vertical velocity is zero because the diver is diving straight out from the cliff horizontally. The only force acting on the diver in the vertical direction is gravity. We can use the equation of motion to find the time it takes for the diver to fall from the cliff to the rocks below.
The equation for vertical motion is:
h = ut + (1/2)gt^2
Where:
h = height (21 m in this case)
u = initial vertical velocity (zero)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Since the diver starts from rest, the equation simplifies to:
h = (1/2)gt^2
Substituting the values, we get:
21 = (1/2)(9.8)t^2
Simplifying further:
21 = 4.9t^2
Dividing by 4.9:
t^2 = 4.2857
Taking the square root:
t ≈ 2.07 seconds
Now that we have the time taken for the diver to fall vertically, let's move on to the horizontal motion.
The horizontal distance traveled by the diver can be calculated using the equation:
d = vt
Where:
d = distance (15 m in this case)
v = horizontal velocity (what we want to find)
t = time (2.07 seconds)
Substituting the known values in the equation:
15 = v * 2.07
Solving for v:
v ≈ 7.25 m/s
Therefore, the diver must have a minimum horizontal velocity of approximately 7.25 m/s to clear the rocks safely when diving straight out from a cliff that is 21 m high and the rocks extend 15 m outward from the cliff.