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lapo4ka [179]
3 years ago
9

What is projectile motion​

Physics
2 answers:
andrezito [222]3 years ago
6 0

Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path.

The path that the object follows is called its trajectory.

Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity

Radda [10]3 years ago
5 0

\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}

\sf{\qquad{\qquad{\underline{\underline{ Projectile~motion }}}}}

If an object is given an initial velocity in any direction and then allowed to travel freely under gravity only, it is called a projectile motion.

It is basically 3 types

  1. horizontally projectile motion
  2. oblique projectile motion
  3. included plane projectile motion

The path followed by a projectile is called its trajectory.

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Answer:

r = 2.84 \times 10^{-14} m

Explanation:

As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle

So here we can write it as

\frac{1}{2}mv^2 = \frac{k(2e)(ze)}{r}

now we know that

m = 1.67 \times 10^{-27} kg

e = 1.6 \times 10^{-19} C

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here kinetic energy of the incident alpha particle is given as

KE = 6.4 \times 10^{-13} J

now we have

6.4 \times 10^{-13} = \frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r}

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r = 2.84 \times 10^{-14} m

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A) When the plane is flying straight down, there are three forces acting on it:
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- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
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(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
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\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
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