a) 5.0 m/s
This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:
[tex]E_i=\frac{1}{2}m_1v_1^2[/tex]
where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.
When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:
[tex]E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh[/tex]
where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find
[tex]\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s[/tex]
b) 4.0 m/s
After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:
[tex]p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3[/tex]
where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find
[tex]v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s[/tex]
c) 2.8 m
We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:
[tex]E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J[/tex]
While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:
[tex]E_f = (m_1 +m_2)gh_{max}[/tex]
where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find
[tex]E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m[/tex]
We have that for the Question ".a)What is the speed of the girl immediately before she grabs the box?b)What is the speed of the girl immediately after she grabs the box?c)What is the maximum height that the girl (with box) reaches?"
it can be said that
From the question we are told
A girl of mass m1=60 kg springs from a trampoline with an initial upward velocity of v1=8.0 m/s. At height h=2.0 m above the trampoline, the girl grabs a box of mass m2= 15 kilograms. using g=9.8 m/s
The equation for the speed of the girl immediately before she grabs the box is mathematically given as
[tex]v^2 = u^2 - 2gs\\\\v^2 = (8)^2 - 2 * 9.8 * 2\\\\v = \sqrt{24.8} \\\\= 4.98 m/s[/tex]
The equation for the speed of the girl immediately after she grabs the box is given as
[tex]= m_1u_1 + m_2u_2 = (m_1+m_2) * v\\\\= 60 * 4.98 + 0 = (60+15) * v\\\\= v = 3.98 m/s[/tex]
The equation for the maximum height that the girl reaches with the box
[tex]v^2 = u^2 - 2gs\\\\= 0 = (3.98)^2 - 2 * 9.8 * s\\\\= s = 0.81m[/tex]
Therefore, total height with respect to the top of the trampoline
[tex]= 2+0.81 = 2.81m[/tex]
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