Answer:
a) s = 5619.2 m
b) v = 456.5 m/s
Note: The preceding question is given below:
An airplane flying horizontally with a speed of 500 km/h at a height of 800 m drops a crate of supplies (see the following figure). If the parachute fails to open, how far in front of the release point does the crate hit the ground?
Explanation:
From the preceding question;
height, h = 800 m,
speed of plane = 500 km/h = 500000 m / 3600 s = 139 m/s
Speed of projectile = 139 m/s + 300 m/s = 439 m/s
a) using h = ut + 1/2 * gt² = 1/2 gt² (sice u = 0)
t = √(2h/g) where g = 9.8 m/s²
t = √(2 * 800/9.8)
t = 12. 8 s
horizontal distance, s = horizontal velocity * time
s = 439 * 12.8
s = 5619.2 m
b) Final vertical velocity v₂ = u - gt
v₂ = 0 - 9.8 * 12.8 = 125.4 m/s
horizontal velocity, vₓ = 439.0 m/s
resultant velocity, v = √(v₁² + vₓ²)
v = √{(125.4)² + (439.0)²}
v = 456.5 m/s
