Answer:
The mailbag will take 2.44 seconds to reach the ground.
Explanation:
The height of a helicopter above the ground is given by:
Height of helicopter at t = 2.10 seconds
The helicopter releases a small mailbag from the height of 29.17 m.
The initial velocity of mailbag = u = 0 m/s
Duration in which mailbag will reach the ground = T
Acceleration due to gravity = g =
Using second equation of motion ;
We have , s = 29.17
u = 0 m/s
t = T
Solving for T, we gte :
T = 2.44 seconds
The mailbag will take 2.44 seconds to reach the ground.
Magnitude is the word used to describe the amount of energy released by an earthquake.
Hope I helped! :D
Radio Waves :)) i’m pretty confident in that
Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m