A 400g lemming waddles horizontally...

[bully3000]

A 400g lemming waddles horizontally off a small 4.9m high cliff with a horizontal velocity of 0.5m/s.
a) How far from the base of the cliff will the lemming land?
b) What will be the lemming's speed just before landing?
c) What is the lemming's kinetic energy just before landing?

[kaiipoo_]

A 400g lemming waddles horizontally off a small 4.9m high cliff with a horizontal velocity of 0.5m/s.
a) How far from the base of the cliff will the lemming land?
b) What will be the lemming's speed just before landing?
c) What is the lemming's kinetic energy just before landing?


This is a projectile motion question. As with almost all projectile motions, what you have to do is break it into it's horizontal and vertical components. you'd know from vectors, that the ACTUAL velocity/acceleration is made up of it's horizontal and vertical vector components.

a) This is basically asking you for the final horizontal displacement. What you need to do in this case, is solve for time using vertical components, which i'll show you how to do.

VERTICALLY:

a= -9.8 m/s² (this is a negative value because in projectile motion, the acceleration due to gravity acts down. It can also be positive, dependent on what direction you take as positive/negative. In this case, i take negative as down)
s= -4.9 m ( as displacement is downwards)
u= 0 m/s ( as it is solely travelling horizontally, meaning there is no upwards or downwards movement)
v= ?
t= ?

We dont want v, we want t. So using an appropriate formula, in this case, s= ut + 1/2at²

Thus, -4.9 = 0.5 x -9.8 x t²
Solving for t, t = 1s. Thus, the entire trip takes 1s.


For the horizontal displacement, understand that there is a negligible air resistance force, which is usually ignored. Thus, the velocity tends to stay constant horizontally. Simply put, we use the nice and easy formula x = vt

Thus the horizontal displacement is 0.5x1 = 0.5 meters.

b) for the final speed, we have to use vectors.
Using the vertical setup above, what we want to find is v.

Thus, we can use the formula v= u + at, or v²-u²= 2as

Usually what you would prefer to do is use the latter, in case you f**ked up part a).

Therefore solving for v, v = 9.8 m/s (the square root of 96)

v horizontally is still 0.5 m/s.

Therefore, Speed_final = sqr root of 9.8² and 0.5²


c) For the Kinetic Energy, simply plug in the above value (final speed) into the formula Ek = 0.5mv²


voila

[lzxnl]

I would personally solve parts b and c using conservation of energy as this way, if your answer to a was wrong, you wouldn't lose further marks. Also, conservation of energy is simpler than vector magnitudes.