Vertical problem:
v = a t
x = Xi + Vi t + .5 a t^2
here
x = 0
Xi = 49.1
Vi = 0
a = -9.8
0 = 49.1 + 0 t - 4.9 t^2
so
a) t = sqrt ( 49.1 m/4.9 m/s^2 )
b) v = -9.8 t
c)
u = 12.6
v = -9.8 t
tan(angle from horizontal) =(v/u)
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 12.6 m/s. The cliff is 49.1 m above a flat horizontal beach as shown in the figure.
a.How long after being released does the stone strike the beach below the cliff?
b.With what speed does the stone land?
c.What is the angle of impact at landing?
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