Question

An airplane flying horizontally with a speed of 500 km/h at a height of 850 m fires a projectile horizontally in its direction of motion at a speed of 290 m/s relative to the plane. (Calculate the following answers relative to the ground.) (a) How far in front of the release point does the projectile hit the ground (in m)? (b) What is its speed when it hits the ground (in m/s)?

Answer 1

The projectile hit the ground 3,992.97 m in front of the release point.  The projectile speed when it hits the ground is 438.5 m/s

When an airplane flying horizontally fires a projectile horizontally in its direction of motion it means

• v₁ = speed of an airplane = 500 km/h
  v₁ = $500 \times \frac{1,000 \: km}{3,600 \: s}$ = 138.89 m/s
• v₂ = speed of a projectile relative to the plane = 290 m/s
• v = speed of a projectile relative to the ground
  v = v₁ + v₂ = 428.89 m/s

The projectile moves in horizontally but it has projectile motion that move horizontally and vertically according to the parabolic curve.

In vertical, the projectile moves in a non-uniform motion.

• uy = the initial speed in vertically = 0 m/s
• h = the height = 850 m
• g = acceleration due to gravity = 9.8 m/s²

• $h = \frac{1}{2} gt^2$
  2h = gt²
  2 × 850 = 9.8 × t²
  $t = \sqrt{1,700}{9.8}$
  t = 9.31 s

Horizontal, the projectile moves in uniform motion.

• Distance
  d = vt
  d = 428.89 × 9.31
  d = 3,992.97 m

When the projectile almost hits the ground

• In vertical
  Final speed
  vy = gt = 9.8 × 9.31 = 91.24 m/s
• In horizontal
  vx = v = 428.89 m/s

Final speed for the projectile

v² = vx² + vy²

v² = 428.89² + 91.24²

v² = 183,946.63 + 8,324.37

v² = 192,271.004

$v = \sqrt{192,271.004}$

v = 438.5 m/s

Learn more about Projectile motion here: https://brainly.com/question/16992646

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