For constant fore the impulse could be increased by increasing the time for the force.
Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
This is a concept of momentum. In equation, momentum is the product of force and distance. When a ball is thrown, its force is constant all throughout unless disturbed by an external force. Therefore, force is the constant of proportionality that relates momentum with distance. When you block a ball from a given distance, you would feel the great force on your hand. In order to reduce the force, you have to follow the direction of the force in order to minimize the impact. By doing this, you gradually decrease the momentum of the ball.
U=1/2kx2
This image sums it up
First we gotta use an equation of motion:
Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!
Now we just need to solve for t:
Hit the calculators, and you'll get 4.5 seconds!
