The height of the bullet when the velocity is zero is 256 ft.
<h3>Height of the bullet when the velocity is zero </h3>
The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;
The height of the bullet at this time is calculated as follows;
Learn more about height of projectiles here: brainly.com/question/10008919
Answer:
Explanation:
We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for
λ
:
λ
=
v
f
Let's plug in our given values and see what we get!
λ
=
340
m
s
440
s
−
1
λ
=
0.773
m
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.
Answer:
At which point does the planet have the least gravitational force acting on it?
Explanation:
In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let 
be the masses of the three fragments.
Let 
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,
So the x- component of the velocity of the m2 fragment after the explosion is,
∴ 
