# MCQ thats giving me a headache

1. Apr 15, 2006

### O.J.

hello,

i have this question thats buggin me :

two markers M1 and M2 are vertically h meters apart. a steel ball is released x m above M1 and reaches M1 by time t1 and M2 by time t2. whats the expression for acceleration?

a.) 2h/(t2)^2
b.) 2h/(t1+t2)
c.) 2h/(t2-t1)^2
d.) 2h/(t2^2-t1^2)

2. Apr 15, 2006

### Andrew Mason

You have to start with the equation for distance fallen. You should know that.

Write out the expression for the distance fallen in time t1 (= x). Write out the expression for the total distance fallen at time t2 (= x+h). Subtract the two equations to get the equation for the difference (h). Work out a from that (which, if it is on earth, is g).

AM

3. Apr 16, 2006

### O.J.

i really am confused. can u show me how to work it out????

4. Apr 16, 2006

In this question, your ultimate goal is to find the acceleration.

As Andrew Mason mentioned, you need to "write out the expression for the distance fallen in time t1 (= x)". What variables are present? Do you know which Kinematics equation you should use?

Similarly, you will need to "write out the expression for the total distance fallen at time t2 (= x+h)". Use the appropriate Kinematics equation here as well.

Carry out some algebraic manipulation, and you should get the answer!

If you need more help, please show some of your working first.

Hint: The steel ball was released from rest.

All the best!

Last edited: Apr 16, 2006
5. Apr 16, 2006

### O.J.

ok, this is my workin,

the vertical distance between the M1 and M2 is h. initial velocity (u) at M1 is found by:

s=(v+u)/2 x t
x=(0+v)/s x t1

we get v=2x / t1, and this is = u for the 2nd part

similarly, v= (2x+2h)/t2

we can also get t as (t2-t1)

so we have v,u,s and t. we can use anythree of them in a kinematics equation to simplify an expression for a. but when im doin it im getting complicated expressions that dont simplify to any of the options

6. Apr 16, 2006

### Andrew Mason

You can use $s = vt/2$ but since v = at, this works out to:

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

so:
(1) $$x = \frac{1}{2}at_1^2$$ and

(2) $$x + h = \frac{1}{2}at_2^2$$

Subtracting (1) from (2):

$$h = \frac{1}{2}at_2^2 - \frac{1}{2}at_1^2$$