A spring-loaded toy dart gun is used to shoot a dart straight up in the air, and the dart reaches a maximum height of 24 m. The same dart is shot straight up a second time from the same gun, but this time the spring is compressed only half as far as before firing. How far up does the dart go this time, neglecting friction and assuming an ideal spring?

 Solution:

Formula used , $\frac{1}{2}kx^2 = mgh$

k = spring constant

$x_{c}$ = spring compressed

h = height

g = acceleration due to gravity

m = mass of the object 

In first scenario,

h = 24 m        

$x_{c}$ = x m

$\frac{1}{2}kx^2$ = mg(24)                 ..... (1)

In second scenario,

$x_{c} = \frac{x}{2}$ m

h = ?

$\frac{1}{2}k (\frac{x}{2})^2$ = mgh                ....(2)

$\frac{1}{2}k \frac{x^2}{4}$ = mgh

$\frac{mg(24)}{4}$ = mgh            [ from (1) ]

$\frac{24}{4}$ = h

h = 6 m