A diver launches off of a 25 m cliff with a speed of 5 m/s and an angle of 30° from the horizontal. How long does it take the diver to hit the water?

1 Answer
Feb 22, 2016

Assuming 30^o is taken below the horizontal
t~=2.0 s.
Assuming 30^o is taken above the horizontal
t~=2.5 s.

Explanation:

Once you know the initial velocity in the y, you can treat this as one dimensional motion (in the y) and ignore the x motion (you only need the x if you want to know how far from the cliff they will land).

Note: I will be treating UP as negative and DOWN as positive for the WHOLE problem.

-Need to know if it's 30^o above or below the horizontal (you probably have a picture)

A) Assuming 30^o below the horizontal, (she jumps down).

We break up the initial velocity of 5 m/s as follows:
v_y = 5 * sin (30^o) m / s [down]
v_y = 5 * 0.5 m / s [down]

v_y = + 2.5 m / s
Note that v_x = 5 * cos (30^o) m / s [away from cliff] ,
but this does NOT affect the answer.

We have the initial velocity v_1 or v_o=2.5 m/sin the y,
the acceleration, a, in the y (just gravity a = 9.8 m/s^2),
the displacement, d=25 m, in the y and want the time, t.
The Kinematic equation that has these terms is given by:
d=v_1 t +1/2 at^2
Subbing in we have
25 m=2.5 m/s t +1/2 9.80 m/s^2 t^2, dropping the units for convince and rearranging we have

0=4.90 t^2 + 2.5 t-25
Put this though the quadratic formula to solve for t.
t_1=-2.5282315724434 and
t_2=2.0180274908108.
In this case the negative root is nonsense, so t~=2.0 s.

B) Assuming 30^o above the horizontal, (she jumps up).

We break up the initial velocity of 5 m/s as follows:
v_y = 5 * sin (30^o) m / s [up]

v_y = 5 * 0.5 m / s [up]

v_y = - 2.5 m / s (positive is down and negative is up!)
Note that v_x = 5 * cos (30^o) m / s [away from cliff] , but this does NOT affect the answer.

We have the initial velocity v_1 or v_o=-2.5 m/sin the y, the acceleration,a, in the y (just gravity a = 9.8 m/s^2), the displacement, d=25 m, in the y and want the time, t. The Kinematic equation that has these terms is given by:
d=v_1 t +1/2 at^2
Subbing in we have
25 m=-2.5 m/s t +1/2 9.80 m/s^2 t^2, dropping the units for convince and rearranging we have

0=4.90 t^2 - 2.5 t-25
Put this though the quadratic formula to solve for t.
t_1=-2.0180274908108 and
t_2=2.5282315724434 (look they switched!)

Again, the negative root is nonsense, so t~=2.5 s.