Homework (Initial velocity, gravity, and falling height)

[RawTalent]

Hey I am having a bit of trouble finding the proper equation to use when solving a problem such as this...

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor at 0.350 s. Ignore air resistance. Find (a) the height of the tabletop above the floor, (b) the horizontal distance from the edge of the table to the point where the book strikes the floor, and (c) the horizontal and vertical components of the book’s velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

I used this equation with no success...
1.10(.35)+1/2(-9.8)(.1225)
= .385 + (-.60025)
= -.215 m

where do i go from here? Thanks.

 

[rl.bhat]

Hi Raw Talent, welcome to PF.
When the book leaves the table top, its velocity in the downward direction is zero and in the horizontal direction is 1.1 m/s, which remains costant.

 

[RawTalent]

So substituting 0 m/s for initial velocity I end up with .6 meters. I am still unsure if that's correct. Can anybody tell me if I'm using the correct equation?

 

[rl.bhat]

Use h = 1/2*g*t^2 and find the answer.

 

[rwisz]

Dear student,

Thank you for reaching out for help with this physics problem. To solve this problem, we can use the equations for motion with constant acceleration. These equations are:

- vf = vi + at (final velocity equals initial velocity plus acceleration multiplied by time)
- d = vit + 1/2at^2 (distance equals initial velocity multiplied by time plus one-half acceleration multiplied by time squared)
- vf^2 = vi^2 + 2ad (final velocity squared equals initial velocity squared plus two times acceleration multiplied by distance)

We can also use the equation for gravitational acceleration, which is g = 9.8 m/s^2.

Now, let's break down the problem and use these equations to find the answers:

(a) The height of the tabletop above the floor can be found using the second equation listed above. We know that the initial velocity (vi) is 1.10 m/s, the time (t) is 0.350 s, and the acceleration (a) is -9.8 m/s^2 (negative because it is acting in the opposite direction of the book's motion). Plugging these values into the equation, we get:

d = (1.10)(0.350) + 1/2(-9.8)(0.350)^2
= 0.385 - 0.6025
= -0.2175 m

Since the book starts and ends at the same height, the height of the tabletop is 0.2175 m above the floor.

(b) The horizontal distance from the edge of the table to the point where the book strikes the floor can be found using the first equation listed above. We know that the initial velocity (vi) is 1.10 m/s, the time (t) is 0.350 s, and the acceleration (a) is 0 m/s^2 (since there is no acceleration in the horizontal direction). Plugging these values into the equation, we get:

d = (1.10)(0.350)
= 0.385 m

Therefore, the horizontal distance from the edge of the table to the point where the book strikes the floor is 0.385 m.

(c) To find the horizontal and vertical components of the book's velocity just before it reaches the floor, we can use the third equation listed above. We know that the final velocity (vf) is 0 m/s (