Why Do Horizontal Motion Problems in AS Physics Challenge Students?

[influx]

I'm currently studying Edexcel AS Physics (a UK-based examboard), I'm comfortable with problems where there is no horizontal component (so when something is thrown/released in a straight line; e.g. a ball thrown upwards), but I really struggle with problems that include the horizontal direction. I know that air resistance is neglected and thus the object will move with a constant velocity but it still doesn't make sense. For example, the below question:

A table tennis ball is hit, without any spin, from one end of a table so that it leaves
the bat horizontally with a speed of 31 m/s. The length of the table is 2.7 m. Show that the ball falls a vertical distance of about 4 cm as it travels the length of the table.

I usually use the UVATS/SUVATS method, but I am unsure what he values of these are in this case?

u = is it 31 m/s ?
v = ?
a = (do we consider this)
t = ?
s = (not sure?)

 

[gneill]

Since you know that the horizontal and vertical components of the velocity are independent of each other, start by listing the known parameters of each. For example, what is the initial velocity in the horizontal direction? How about the vertical direction?

If you consider the horizontal motion alone, how long does it take to complete the portion of its journey that is of interest?

 

[influx]

Since you know that the horizontal and vertical components of the velocity are independent of each other, start by listing the known parameters of each. For example, what is the initial velocity in the horizontal direction? How about the vertical direction?

If you consider the horizontal motion alone, how long does it take to complete the portion of its journey that is of interest?

I think I got it:

Horizontal:

u = 31 m/s
s = 2.7 m

since s = ut
s/u = t = 2.7/31 = 0.0870 seconds

Vertical:

u = 0 m/s
t= 0.0870 s
a= 9.8 m/s²
s = s

s = 0.5(9.8)(0.0870)²
s = 0.0371 m
s= 3.71 cm

The time a projectile is in flight, is always the same when considering vertical and horizontal right?

Cheers

 

[gneill]

I think I got it:

Horizontal:

u = 31 m/s
s = 2.7 m

since s = ut
s/u = t = 2.7/31 = 0.0870 seconds

Vertical:

u = 0 m/s
t= 0.0870 s
a= 9.8 m/s²
s = s

s = 0.5(9.8)(0.0870)²
s = 0.0371 m
s= 3.71 cm

Yes, your method and results look good. Nicely done.

The time a projectile is in flight, is always the same when considering vertical and horizontal right?

Right. It couldn't be otherwise unless the projectile could be in two places at once!

 

[influx]

Since you know that the horizontal and vertical components of the velocity are independent of each other, start by listing the known parameters of each. For example, what is the initial velocity in the horizontal direction? How about the vertical direction?

If you consider the horizontal motion alone, how long does it take to complete the portion of its journey that is of interest?

Yes, your method and results look good. Nicely done.

Right. It couldn't be otherwise unless the projectile could be in two places at once!

Ahh yes! Thanks :)