Question

A tiger leaps horizontally from a 6.1 m high rock with a speed of 3.9 m/s. How far from the base of the rock will she land?

Answer 1

The horizontal distance (s)= ut+
`(1)/(2) g t^(2)`
Time taken by the ball to hit the ground,
h=
`(1)/(2) g t^(2)`
h=60
60=
`(1)/(2)*10* t^(2)`
60=5
`t^(2)`

`t^(2)= (60)/(5)`
t=
`√(12)`
Replacing for distance covered
H=ut+
`(1)/(2) g t^(2)`
H=3.9m/s*
`√(12) + (1)/(2) *10* √(12)^(2)`
H=13.51+60
=73.51 m

Answer 2

v o = 3.9 m/s
x = v o * t
h = g t² / 2
6.1 m = 9.81 m/s² * t² /2
t² = 12.1 / 9.81
t = √12.1/9.81
t = 1.11 s
x = 3.9 m/s * 1.11 s
x = 4.329 m
A tiger will land 4.329 m from the base of the rock.