Atiger leaps horizontally from a 12 m high rock with a speed of 4.5 m/s. how far from the base of the rock will she land?

7.02 m away from the base of the rock.

Explanation:

As the tiger leaps from 12 m high rock so, its initial velocity will be zero. According to second equation of motion:

h = (Vi)(t)+½ gt^2

12 = (0)(t)+1/2(9.8)t^2

t^2 = (12)(2)/(9.8)

t = √2.44

t = 1.56 s

Now,  

X = (V)(t)

where,

V = average velocity = 4.5 m/s

X = (4.5)(1.56)

X = 7.02 m

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