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

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.

Answer 2

The horizontal distance (s)= ut+$\frac{1}{2} g t^{2}$
Time taken by the ball to hit the ground,
h=$\frac{1}{2} g t^{2}$
h=60
60= $\frac{1}{2}*10* t^{2}$
60=5$t^{2}$
$t^{2}= \frac{60}{5}$
t=$\sqrt{12}$
Replacing for distance covered
H=ut+$\frac{1}{2} g t^{2}$
H=3.9m/s*$\sqrt{12} + \frac{1}{2} *10* \sqrt{12}^{2}$
H=13.51+60
=73.51 m