Question

A lunch pail is accidentally kicked off a steel beam on a building under construction. Suppose the initial horizontal speed is 1.50 m/s. How far does the lunch pail fall after it travels 3.50 m horizontally? 8. If the building in problem 7 is 2.50 × 10 2 m tall, and the lunch pail is knocked off the top floor, what will be the horizontal displacement of the lunch pail when it reaches the ground?

Answer 1

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:

$d=v_x t$

where we know

$v_x = 1.50 m/s$ is the horizontal velocity

d = 3.50 m is the distance covered horizontally

Solving for t, we find the total time of the motion:

$t=\frac{d}{v_x}=\frac{3.50}{1.50}=2.33 s$

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

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Substutting t = 2.33 s, we find how fat the pail has fallen:

$s=\frac{1}{2}(9.8)(2.33)^2=26.6 m$

2) 10.7 m

In this case, we know instead the vertical displacement:

$s=2.50\cdot 10^2 m = 250 m$

Therefore, we can use the same equation again

$s=ut+\frac{1}{2}at^2$

To find the total time of motion:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(250)}{9.8}}=7.14 s$

We know that along the horizontal direction, the velocity is constant:

$v_x = 1.50 m/s$

So, the horizontal distance covered in this time is

$d=v_x t = (1.50)(7.14)=10.7 m$