Answer:
The maximum height above its initial position is:
[tex]h_{max}=1.53\: m[/tex]
Explanation:
Using momentum conservation:
[tex]m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb}[/tex] (1)
Where:
Let's find v(fb) using equation (1)
[tex]m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}[/tex]
[tex]v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}[/tex]
[tex]v_{fB}=\frac{0.1(900-300)}{2}[/tex]
[tex]v_{fB}=30\: m/s[/tex]
We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.
[tex]\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}[/tex]
[tex]h_{max}=\frac{1}{2g}v_{fB}[/tex]
[tex]h_{max}=\frac{1}{2(9.81)}30[/tex]
[tex]h_{max}=1.53\: m[/tex]
I hope it helps you!
The maximum height above its initial position is: [tex]h_{max}[/tex]=1.53m
Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
Using momentum conservation:
[tex]m_bv_{1b}=m_Bv_{fB}+m_bv_{fb}[/tex]
Where:
m(b) is the mass of the bullet
m(B) is the mass of the block
v(ib) is the initial velocity of the bullet
v(fb) is the final velocity of the bullet
v(fB) is the final velocity of the block
Now for finding the value of final velocity VfB
[tex]m_b(v_{1b}-v_{fb})=m_Bv_{fB}[/tex]
[tex]v_{fB}=\dfrac{m_b(v_{1b}-v_{fb})}{m_B}[/tex]
[tex]v_{fB}=\dfrac{0.1(900-{300})}{2}[/tex]
[tex]v_{fB}=30\ \frac{m}{s}[/tex]
We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.
[tex]\dfrac{1}{2}m_Bv_{fB}=m_Bv_{fB}[/tex]
[tex]h_{max}=\dfrac{v_{fB}}{2g}[/tex]
[tex]h_{max}=\dfrac{30}{2\times 9.81}[/tex]
[tex]h_{max}=1.53\ m[/tex]
Thus the maximum height above its initial position is: [tex]h_{max}[/tex]=1.53m
To know more about Simple harmonic motion, follow
https://brainly.com/question/17315536
