Asmall wooden block with mass 0.725 kg is suspended from the lower end of a light cord that is 1.68 m long. the block is initially at rest. a bullet with mass 0.0110 kg is fired at the block with a horizontal velocity v0. the bullet strikes the block and becomes embedded in it. after the collision the combined object swings on the end of the cord. when the block has risen a vertical height of 0.700 m, the tension in the cord is 5.00 n. what was the initial speed v0 of the bullet?

vo =  277.7m / s

Explanation:

We will solve this problem in two parts, the first one as an inelastic shock where we will find the speed of the most bullet block system and the second with the conservation of the system's energy

We create a system formed by the bullet and block assembly, in this system the forces of the shock are internal and the moment is preserved, call us to the mass of the bullet (m) and the mass of the block (M)

before the crash

     p₀ = m v₀

just after the crash

     pf = (m + M) vf

Like the inelastic shock

    p₀ = pf

    m v₀= (m + M) vf

    vf = m / (m + M) v₀               (1)

Now we use energy conservation for this system, let's write mechanical energy at two points; an initial point at the lowest part and an end point at the height part 0.700m

Initial

     Em₀ = K = ½ (m + M) vf²

Final

We do not know if this is the highest point of the trajectory

     Em₂ = U + K = mgy + ½ (m + M) v²

     Em₀ = Em₂

     ½ (m + M) vf² = (m + M) g + ½ (m + M) v²          (2)

The movement of the system is circulated, so with Newton's second law we can find the speed at this point

     sin θ = Tₓ / T

     cos θ = Ty / T

     Tₓ = T sin θ

     Ty = T cos θ

Y Axis

    Ty -W = 0

X axis

    Tₓ = m a

Where the acceleration is centripetal

    a = v² / r

    T sin θ = m v² / r

    v² = T r sin θ

Let's calculate

   sin θ = y / L

   sin θ = 0.700 / 1.68

   sin θ = 0.416

   θ = 24.6º

   v² = 5 1.68 0.416

   v² = 3.4944 m2 / s2

   v  = 1.869 m/s

We substitute in equation 2

   ½ (m + M) vf² = (m + M) g y + ½ (m + M) 1.869²  

             vf² = 2 g y + 3,494

             vf² = 2 9.8 0.700 + 3.494

             vf² = 17.214 m/s

             vf = 4.15 m/s

This is the speed with which the system leaves at the lowest point of the path

Now we substitute in equation 1 and calculate the initial velocity (vo)

            vf = m/(m + M)  vo

            vo = vf (m + M)/m

            vo = 4.15 (0.0110 + 0.725) / 0.0110

            vo =  277.7m / s

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .