Question

Please look at the answers below before solving!

B x (A) (8 m 32 m 4テitetimum. Sme time during its f ig, it ex,lodes into two fragineds, A and B, of maem1o a1.5 respecfively. And exactly 6 s offer the initial frng u fragmets land meousy af fh indicated pin, Find the e ifil velbecif oud tnt

Answer:

a) v0 = 29.7 m/s

Theta = 81.5 degrees

Answer 1

x_cm = position of center of mass of the fragments

x_A = position of first fragment relative to launch point = 18 m

m_A = mass of first fragment = 1 kg

x_B = position of second fragment relative to launch point = 32 m

m_B = mass of second fragment = 1.5 kg

Location of center of mass is given as

x_cm = (m_A x_A + m_B x_B)/(m_A + m_B) = ((1) (18) + (1.5) (32))/(1 + 1.5) = 26.4 m

X = horizontal range of the projectile when no explosion take place = x_cm = 26.4 m

t = time of flight = 6 sec

Consider the motion along the horizontal direction :

X = horizontal range of the projectile when no explosion take place = x_cm = 26.4 m

t = time of flight = 6 sec

v_ox = component of velocity along x-direction

using the equation

X = v_ox t (since no acceleration along x-direction)

26.4 = 6 v_ox

v_ox = 4.4 m/s

Consider the motion along the vertical direction :

Y = displacement = 0 m

v_oy = component of velocity along y-direction

a_y = acceleration = - 9.8 m/s²

t = time of flight = 6 sec

using the equation

Y = v_oy t + (0.5) a_y t²

0 = v_oy (6) + (0.5) (- 9.8) (6)²

v_oy = 29.4 m/s

speed of launch is given using Pythagorean theorem as

v_o = sqrt(v_ox² + v_oy²) = sqrt((4.4)² + (29.4)²) = 29.73 m/s

angle is given as

$\theta$ = tan^-1(v_oy/v_ox) = tan^-1(29.4/4.4) = 81.5 deg