2. Above right: The four positions of the thrown ball with no gravity are
1 answer:
The characteristics of the projectile launch allows to find the results for the questions about the movement of the ball are:
- In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.
- In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.
Kinematics studies the motion of bodies looking for relationships between position, velocity and acceleration. In the case of vertical and projectiles launch the acceleration on the vertical axis is the acceleration of gravity directed downward.
In the attachment we can see the position of the ball for two distances in the case of projectile launching.
We can see that the speed of the ball decreases with height according to the relation
y = go t - ½ g t²
Where y is the height, g is the initial vertical velocity, g is the acceleration of gravity and t is time.
In all movements in various dimensions we assume that each movement in an . axis is independent
In the case of projectile launching, on the vertical axis there is an acceleration of gravity and on the horizontal axis there is no acceleration, the only parameter that this gives the two movements is the time, which is a scalar.
In conclusion, using the characteristics of the projectile launch, we can find the results for the questions about the movement of the ball are:
- In the movement is several dimensions, each one is independent of the others, the movement in the x axis does not affect the movement in the y axis.
- In the attached we see the vertical speed decreases with height and the curve is of the parabolic type.
Learn more about projectile launch here: brainly.com/question/24888457
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Answer:
The minimum distance in which the car will stop is
x=167.38m
Explanation:
∑F=m*a
∑F=u*m*g
The force of friction is the same value but in different direction of the force moving the car so it can stop so
The coefficient of friction is missing and it has a value of μ = 0.4
Answer:
a = 3.924 m/s²
Explanation:
I've attached the kinematic free body diagram.
Taking the sum of all upward and downward forces,
EFy = 0;
N1 + N2 - m_p•g - m_v•g = 0
N1 + N2 = m_p•g + m_v•g
Where;
N1 and N2 are the normal reactions at the wheels
m_p is the mass of the driver
m_v is the mass of the vehicle
g is the acceleration due to gravity.
Plugging in the relevant values in the question,we obtain;
N1 + N2 = (385 + 75) x 9.81
N1 + N2 = 4512.6N - - - (eq1)
Now, taking sum of all horizontal forces;
EFx = (m_p + m_v) x a
So,
μ(N1 + N2) = (mp + mv) x a
Thus,
0.4(N1 + N2) = (385 + 75)a
0.4(N1 + N2) = 460a
N1 + N2 = 1150a
From eq(1),N1 + N2 = 4512.6N
Thus,
1150a = 4512.6N
a = 4512.6/1150
a = 3.924 m/s²
Therefore, the acceleration, a = 3.924 m/s²
