A person dives off the edge of a cliff 40m above the surface of the sea below. Assuming that air resistance is negligible, (a) How long does it dive last? (b) and with what speed does the person enter the water?
a. d = Vo*t + 0.5at^2 = 40m,
0 + 0.5 * 9.8 * t^2 = 40,
4.9t^2 = 40,
t^2 = 8.16,
t = 2.86s.
b. Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2 * 9.8 *40 = 784,
Vf = 28m/s.
A) h=0.5gt^2
40=0.5(9.8)(t^2)
40=4.9t^2
t^2=8.16
t=2.86s
To solve this problem, we can use the equations of motion under constant acceleration.
(a) To find the time it takes for the person to dive, we can use the equation:
h = (1/2) * g * t^2
where:
h = height (40 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for time:
t = sqrt((2 * h) / g)
Substituting the given values:
t = sqrt((2 * 40) / 9.8)
t ≈ 2.03 seconds
Therefore, the dive lasts approximately 2.03 seconds.
(b) To find the speed with which the person enters the water, we can use the equation:
v = g * t
where:
v = speed
g = acceleration due to gravity (9.8 m/s^2)
t = time
Substituting the value of time we found in part (a):
v = 9.8 * 2.03
v ≈ 19.92 m/s
Therefore, the person enters the water with a speed of approximately 19.92 m/s.