7. A cat pushes a ball from a 7.00 m high window, giving it a horizontal velocity of 0.20 m/s. How far from the base of the building does the ball land?
(Assume no air resistance and that ay= g = 9.81 m/s2.)
H=gt²/2
t=sqrt(2H/g) = sqrt(2•7/9.81)=1.2 s.
x=v(x) •t =0.2•1.2 =2.4 m
.24
To solve this problem, we can use the equations of motion for projectile motion. The vertical and horizontal motions of the ball are independent of each other.
First, let's calculate the time it takes for the ball to reach the ground using the vertical motion equation:
h = ut + (1/2)gt^2
where h is the initial height (7.00 m), u is the initial vertical velocity (0 m/s, since the cat pushes only horizontally), g is the acceleration due to gravity (-9.81 m/s²), and t is the time it takes for the ball to reach the ground.
Plugging in the values:
7 = 0*t + (1/2)(-9.81)t^2
Simplifying:
4.905t^2 = 7
t^2 = 7 / 4.905
t = √(7 / 4.905)
t ≈ 1.12 sec
Now that we know the time it takes for the ball to reach the ground, we can calculate the horizontal distance it travels using the horizontal motion equation:
d = vt
where d is the horizontal distance, v is the initial horizontal velocity (0.20 m/s), and t is the time we just calculated (1.12 sec).
Plugging in the values:
d = 0.20 * 1.12
d ≈ 0.224 meters
Therefore, the ball will land approximately 0.224 meters from the base of the building.