A tiger leaps horizontally from a 5.9 meter high rock with a speed of 3.7 m/s.

Questions: How far from the base of the rock will the tiger land?

2 answers

H=gt²/2 => t =sqrt(2H/g) =sqrt(2•5.9/9.8) =1.1 s.
X =v•t =3.7•1.1 = 4.07 m.
y=1/2at^2|
5.9=1/2(9.8)t^2|
1.204=t^2|
t=1.097s|
___________

x=t[(Vi+Vf)/2]
x=1.097[(3.7+0)/2]
x=2.03m