# A ball with a mass of 350 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16 (kg)/s^2 and was compressed by 7/4 m when the ball was released. How high will the ball go?

Apr 11, 2018

The height reached by the ball is $= 7.14 m$

#### Explanation: The spring constant is $k = 16 k g {s}^{-} 2$

The compression of the spring is $x = \frac{7}{4} m$

The potential energy stored in the spring is

$P E = \frac{1}{2} k {x}^{2} = \frac{1}{2} \cdot 16 \cdot {\left(\frac{7}{4}\right)}^{2} = 24.5 J$

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

$K {E}_{b a l l} = \frac{1}{2} m {u}^{2}$

Let the height of the ball be $= h$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Then ,

The potential energy of the ball is $P {E}_{b a l l} = m g h$

Mass of the ball is $m = 0.350 k g$

$P {E}_{b a l l} = 24.5 = 0.350 \cdot 9.8 \cdot h$

$h = 24.5 \cdot \frac{1}{0.350 \cdot 9.8}$

$= 7.14 m$

The height reached by the ball is $= 7.14 m$

Apr 11, 2018

First we have to find the force of the spring.

#### Explanation:

Using the spring constant formula this can be found

$F = - k x$

$F = 16 \times \frac{7}{4}$

$F = 28 N$

Then the acceleration is:

$a = \frac{F}{m}$

$a = \frac{28}{0.35}$

$a = 80$ $m {s}^{-} 2$

To find the velocity at which the ball leaves the spring the following formula can be used:

${v}^{2} = {u}^{2} + 2 a x$

${v}^{2} = 0 + 2 \times 80 \times \frac{7}{4}$

${v}^{2} = 280$

$v = 16.73$ $m {s}^{-} 1$

Now this is a projectile motion question.

${v}^{2} = {u}^{2} + 2 a H$

$0 = {16.73}^{2} + 2 \times - 9.8 \times H$

$H = \frac{280}{2 \times 9.8}$

$H = 14.29$ $m$

The ball travels $14.29$ $m$ high.

Apr 11, 2018

$\approx 7 m$

#### Explanation:

I will assume that all of the potential energy in the spring will turn into kinetic energy and the only thing opposing the motion of the ball is the acceleration of free fall acting in the opposite direction of it's motion.

Now, the potential energy in the spring is given by:

${E}_{p} = 0.5 k \Delta {x}^{2}$
k=spring constant
x=compression of the spring

${E}_{p} = 0.5 \times \left(16\right) \times {\left(\frac{7}{4}\right)}^{2} = 24.5 J$

${E}_{p} = {E}_{k}$

${E}_{k} = 0.5 \times m \times {v}^{2}$

$24.5 = 0.5 \times 0.35 \times {v}^{2}$

$v = 11.71 m {s}^{-} 1$ (the initial velocity of the ball)

We know that our final velocity will be 0 as this is the point where the ball will start falling down again due to the acceleration of free fall acting in the opposite direction.

Using the kinematic equation:
$v = u + a t$
v=final velocity (0)
u=inital velocity
a=acceleration (-9.81, as gravity is opposing or direction of motion)
t=time

$0 = 11.71 - 9.81 t$
$t = 1.193 s$
So it takes 1.193 seconds for the ball to reach its maximum height.

Now using another kinematic equation to find the distance travelled:

$s = \frac{\left(v + u\right) t}{2}$
s=distance travlled
v=final velocity (0)
u=inital velocity- 11.71
t=time

$s = \frac{1.193 \times 11.71}{2}$
$s = 6.98 m \approx 7 m$