Respuesta :
Given: Initial Velocity Vi = 100 m/s; Altitude y = 1.5 Km; Time t = ?
The aircraft is moving horizontally on x direction
Formula to follow: x = VicosΘt but Time t is unknown.
Solving for Time "t" formula will be y = VisineΘt + 1/2 gt²
Since sineΘt = 0 therefore y = 1/2gt² Solve for t
t = √2y/g t = √2(1500 m)/9.8m/s²) t = 31.24 s
Now solving for distance
x = VicosΘt x = (100 m/s) (31.24 s) x = 3,124 m or 3.124 Km
The aircraft is moving horizontally on x direction
Formula to follow: x = VicosΘt but Time t is unknown.
Solving for Time "t" formula will be y = VisineΘt + 1/2 gt²
Since sineΘt = 0 therefore y = 1/2gt² Solve for t
t = √2y/g t = √2(1500 m)/9.8m/s²) t = 31.24 s
Now solving for distance
x = VicosΘt x = (100 m/s) (31.24 s) x = 3,124 m or 3.124 Km
1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m