Answer:
After leaving the edge, it takes the diver 2.79 s to reach the water.
She travels 21.4 m horizontally from the cliff.
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
The equation of the position vector of the diver at time t is the following:
r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
θ = angle of falling.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s²).
If we place the origin of the frame of reference at the top of the cliff on the edge, then x0 and y0 are equal to zero (initial position r0 = (0, 0)).
First, let's calculate the initial velocity of the diver knowing that she traveled 50.0 m in 6.10 s at constant speed:
v = d/t
where:
v = speed.
x = traveled distance.
t = time.
v = 50.0 m/ 6.10 s
v = 8.20 m/s
The initial velocity is 8.20 m/s.
We know that the vertical component of the position vector at the time when the diver hits the water is -30.0 m (see r1y in the figure). Then, using the equation of vertical position we can calculate the time it takes the diver to reach the water:
y = y0 + v0 · t · sin θ + 1/2 · g · t² (y0 = 0)
-30.0 m = 8.20 m/s · sin (21.0°) · t - 1/2 · 9.8 m/s² · t²
0 = 30.0 m + 8.20 m/s · sin (21.0°) · t - 4.9 m/s² · t²
Solving the quadratic equation:
t = 2.79 s (the other solution is rejected because it is negative).
After leaving the edge, it takes the diver 2.79 s to reach the water.
Now, let's use the equation of the horizontal position and find the horizontal distance traveled in 2.79 s:
x = x0 + v0 · t · cos θ (x0 = 0)
x = 8.20 m/s · 2.79 s · cos (21.0°)
x = 21.4 m
She travels 21.4 m horizontally from the cliff.
