Respuesta :
We need to find the time it took the cat to hit the floor in order to find the cat's speed.
To find this time, we need to consider the vertical component of motion. We can assume the cat slid horizontally off the edge of the table, so the vertical component of the speed is zero. We have enough information to find the time it took the cat to hit the floor using one of the kinematic equations of motion in the vertical direction, taking downwards to be the positive direction.
H = height = 2.5 m
d = horizontal distance = 2.2 m
u = initial velocity = 0 m/s
= acceleration (due to gravity) = 9.8 m/s²
t = time = ?
v = the cat's speed = ?
(I won't rearranged the problem for you)
0.71429s
Now we turn our attention to the horizontal direction. What horizontal forces acted on the cat as it fell? None. So that means no horizontal acceleration and therefore the horizontal speed is constant. So the average horizontal speed is equal to its speed, when it slid off the table, and when speed is constant, horizontal speed is just horizontal distance divided by time
(I can't show the equation as brainly won't let me do complicated problems but i'll help with the answer)
[tex]v=\frac{d}{t}=\frac{2.2m}{0.71429s}=3.07998\:\frac{m}{s}[/tex]
This can be solved by separating the vertical and horizontal motion.
The cat slides off the table and drops with a vertical acceleration of gravity, g, and initial vertical velocity of zero.
Using the kinematic equation: s = ut + 1/2 a t^2,
and substitute s=2.5, u=0, a=g
2.5=1/2gt^2
t = sqrt(5/g)
Assume the cat slides off with a horizontal speed of V, its horizontal acceleration is zero.
Using the same kinematic equation: s = ut + 1/2 a t^2,
and substitute s=2.2, a=0, t=sqrt(5/g)
2.2 = u(sqrt(5/g)) + 1/2 (0) (%/g)
u = 2.2/(sqrt(5/g))
assuming g=9.8, u = 2.2/(sqrt(5/9.8))
u = 3.08 m/s