An Alaskan rescue plane traveling 37 m/s drops a package of emergency rations from a height of 134 m to a stranded party of expl
1 answer:
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At a distance of 469.2 m from the original point below the airplane.
Explanation:
First of all, we have to calculate the time it takes for the sack to reach the ground.
To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:
where, taking downward as positive direction:
s = 300 m is the vertical displacement
u = 0 is the initial vertical velocity
t is the time
is the acceleration of gravity
Solving for t, we find the it takes for the sack to reach the ground:
Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:
We also know the total time of flight,
t = 7.82 s
Therefore, we can find the horizontal distance travelled by the sack:
So, the sack will land 469.2 m from the original point below the airplane.
Learn more about free fall and projectile motion:
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s