Answer:
The helicopter should be more than 13 m away from the launching site.
Explanation:
Hi there!
The equation of velocity of the projectile is the following:
v = v0 + g · t
Where:
v = velocity of the projectile at a time t.
v0 = initial velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
The projectile starts traveling downward after reaching its maximum height. At that point, the velocity of the projectile is zero. Using the equation of velocity, we can find the time at which the projectile is at its maximum height.
The helicopter must be on the path of the projectile after that time to be hit when the projectile is falling.
Let's find the time at which the projectile is at its maximum height (v = 0):
v = v0 + g · t
0 = 28 m/s - 9.8 m/s² · t
-28m/s / -9.8 m/s² = t
t = 2.9 s
Now, let's find the distance (d) that the helicopter travels in that time:
d = v · t
Where "v" is the velocity of the helicopter and "t" is the time.
d = 4.5 m/s · 2.9 s
d = 13 m
If the helicopter is located at 13 m from the launching site, the projectile will hit it at its maximum height. So, the helicopter should be more than 13 m away from the launching site to be hit by the projectile while it is falling.
The helicopter should be more than 13 m away from the launching site.
Velocity of projectile is given by:
[tex]v = v_0 + g * t[/tex]
where:
v = velocity of the projectile at a time t.
v₀ = initial velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
On falling downwards , the velocity of the projectile is zero.
Using the equation of velocity, we can find the time at which the projectile is at its maximum height.
The helicopter must be on the path of the projectile after that time to be hit when the projectile is falling.
Let's find the time at which the projectile is at its maximum height (v = 0):
[tex]v = v_0 + g * t\\\\0 = 28 m/s - 9.8 m/s^2 * t\\\\t=\frac{-28m/s}{-9.8m/s^2} \\\\t = 2.9 s[/tex]
Now, to find the distance that the helicopter travels in that time, we will substitute the values in this formula:
[tex]\text{Velocity}=\frac{\text{Distance}}{\text{Time}}[/tex]
On adding given values to this formula we will get:
[tex]d = 4.5 m/s * 2.9 s\\\\d = 13 m[/tex]
If the helicopter is located at 13 m from the launching site, the projectile will hit it at its maximum height.
So, the helicopter should be more than 13 m away from the launching site to be hit by the projectile while it is falling.
Learn more:
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