A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 830 N/m and the mass of the rocket is 50 g, how high will the rocket fly?

6.86 meters

Explanation:

Let the compression of the string be represented by x, and the height of projection of the toy rocket be represented by h.

So that;

x = 9 cm = 0.09 m

In its rest position (i.e before the launch), the spring has a stored potential energy which is given as;

PE = K

    = x 830 x

    = 415 x 0.0081

    = 3.3615

The potential energy in the string = 3.36 Joules

Also,

PE = mgh

where: m is the mass, g is the gravitational force and h the height.

m = 50 g = 0.05 kg, g = 9.8 m

Thus,

PE = 0.05 x 9.8 x h

3.3615 = 0.05 x 9.8 x h

3.3615 = 0.49h

⇒ h =

      = 6.8602

The height of the toy rocket would be 6.86 meters.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

answer; i believe the correct answer is (40,000 j) good luck