Answer :
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 = [tex]500 \times \frac{1,000 \: km}{3,600 \: s}[/tex] = 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
- [tex]h = \frac{1}{2} gt^2[/tex]
2h = gt²
2 × 850 = 9.8 × t²
[tex]t = \sqrt{1,700}{9.8}[/tex]
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
[tex]v = \sqrt{192,271.004}[/tex]
v = 438.5 m/s
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