first do the vertical problem
We will have to assume she has no initial speed up, just horizontal
h = h initial + Vi t - 4.9 t^2
0 = h initial + 0 (1.2) - 4.9 (1.2)^2
solve for h initial
her horizontal speed never changed so horizontal distance is 2.77*1.
a girl jumps off a high diving platform with a horizontal velocity of 2.77m/s and lands in the water 1.2s later the acceleration of gravity is 9.8m/s squared
how high is the platform and how far from the base of the platform does she land?
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