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/s#in 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/s#in 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#.
