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An m = 6.11 g bullet is fired into a 365 g block that is initially at rest at the edge of a table of h = 1.08 m height as shown in the figure.
1. The bullet remains in the block, and after the impact the block lands d = 1.84 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?
2. What is the initial horizontal velocity of the block as it flies off the table?
3. Determine the initial speed of the bullet.

An m 611 g bullet is fired into a 365 g block that is initially at rest at the edge of a table of h 108 m height as shown in the figure 1 The bullet remains in class=

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onyebuchinnaji

(a) The time taken the block to reach the ground once it flies off the edge of the table is 0.47 s.

(b) The initial horizontal velocity of the block as it flies off the table is 3.91 m/s.

(c) The initial speed of the bullet is 237.5 m/s.

Time of motion of the block

The time of motion of the block is calculated as follows;

h = ¹/₂gt²

where;

  • h is height of the table
  • t is time of motion

t = √(2h/g)

t = √( (2 x 1.08) / 9.8)

t = 0.47 s

Initial horizontal velocity of the block

x = vt

v = x/t

v = (1.84)/(0.47)

v = 3.91 m/s

Initial speed of the bullet

m1u1 + m2u2 = v(m1 + m2)

where;

  • m1 is mass of bullet
  • m2 is mass of block
  • u1 is initial velocity of the bullet
  • u2 is the initial velocity of the block
  • v is initial horizontal velocity of the block as it flies off the table

0.00611(u1) + 0.365(0) = 3.91(0.00611 + 0.365)

0.00611u1 = 1.451

u1 = 1.451/0.00611

u1 = 237.5 m/s

Learn more about horizontal velocity here: https://brainly.com/question/24949996

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