A student drops an object from a fourth floor window the velocity of the objects is measured at 39.28m/s 3 seconds later what is
1 answer:
You might be interested in
This question is incomplete, the complete question is;
Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks; γ is the damping coefficient, with units of kg/sec
m = 0.2, γ = 1.6 and k = 4
Initial displacement is 1 and initial velocity is -2
x" + _____ x' ____x = 0
x(t) =
Answer:
the solution that satisfies the initial conditions specified is;
x(t) = +
Explanation:
Given the data in the question ;
m = 0.2, γ = 1.6, k = 4
x(0) = 1, x'(0) = -2
Now, the differential equation that governs the motions of spring mass system is;
mx" + γx' + kx = 0
so we substitute
0.2x" + 1.6x' + 4x = 0
divide through by 0.2
x" + 8x' + 20x = 0
hence, characteristics equation will be;
m² + 8m + 20 = 0
we find m using; x = [ -b±√(b² - 4ac) ] / 2a
m = [ -8 ± √((8)² - 4(1 × 20 )) ] / 2(1)
m = [ -8 ± √( 64 - 80 ) ] / 2
m = [ -8 ± √-16 ) ] / 2
m = ( -8 ± 4i ) / 2
m = -4 ± 2i
Hence, the general solution of the differential equation is;
x(t) = +
From the initial conditions;
c₁ = 1, c₂ = 1
the solution that satisfies the initial conditions specified is;
x(t) = +
Answer:
The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.
Explanation:
Given;
mass of bullet, m₁ = 4.2 g = 0.0042 kg
mass of hunter + gun = 72.5 kg
velocity of the bullet, u = 965 m/s
Momentum of the bullet when it was fired;
P = mv
P = 0.0042 x 965
P = 4.053 kg.m/s
Determine the recoil velocity of the hunter.
Total momentum = sum of the individual momenta
Total momentum = momentum of the bullet + momentum of the hunter
Apply the principle of conservation of momentum, sum of the momentum is equal to zero.
Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.