1) 26.6 m
Along the horizontal direction, the lunch pail is moving with a uniform motion (constant speed), since there are no forces acting in this direction.
Therefore, the distance travelled horizontally after a time t is given by:
[tex]d=v_x t[/tex]
where we know
[tex]v_x = 1.50 m/s[/tex] is the horizontal velocity
d = 3.50 m is the distance covered horizontally
Solving for t, we find the total time of the motion:
[tex]t=\frac{d}{v_x}=\frac{3.50}{1.50}=2.33 s[/tex]
Now we know that the pail takes 2.33 s to fall to the ground. We can now consider the vertical motion of the pail, which is a free fall motion, so the vertical displacement is given by the equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, taking downward as positive direction:
u = 0 is the initial vertical velocity
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Substutting t = 2.33 s, we find how fat the pail has fallen:
[tex]s=\frac{1}{2}(9.8)(2.33)^2=26.6 m[/tex]
2) 10.7 m
In this case, we know instead the vertical displacement:
[tex]s=2.50\cdot 10^2 m = 250 m[/tex]
Therefore, we can use the same equation again
[tex]s=ut+\frac{1}{2}at^2[/tex]
To find the total time of motion:
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(250)}{9.8}}=7.14 s[/tex]
We know that along the horizontal direction, the velocity is constant:
[tex]v_x = 1.50 m/s[/tex]
So, the horizontal distance covered in this time is
[tex]d=v_x t = (1.50)(7.14)=10.7 m[/tex]
