Alunch 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?

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:

where we know

is the horizontal velocity

d = 3.50 m is the distance covered horizontally

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

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

where, taking downward as positive direction:

u = 0 is the initial vertical velocity

is the acceleration of gravity

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

2) 10.7 m

In this case, we know instead the vertical displacement:

Therefore, we can use the same equation again

To find the total time of motion:

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

So, the horizontal distance covered in this time is

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio