When the projectile is at its maximum height above ground, it's at the point
of changing from rising to falling. At that exact point, its vertical speed is zero,
so the 14 m/s must be all horizontal velocity. That's not going to change.
Since we need to consider changes in vertical speed now, we need to make
some assumption about where this is all happening, so that we know the
acceleration of gravity. I'll assume that it's all happening on or near the Earth,
and the acceleration of gravity is 9.8 m/s².
I'm also going to neglect air resistance.
a). 1.2 sec before it reaches its maximum height, the projectile is rising
at a vertical speed of (1.2 x 9.8) = 11.76 m/s.
The magnitude of its velocity is
the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° above horizontal.
b). 1.2 sec after it reaches its maximum height, the projectile is falling
at a vertical speed of (1.2 x 9.8) = 11.76 m/s.
The magnitude of its velocity is
the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° below horizontal.
===========================
In 1.2 second before or after zero vertical speed, an object in free fall moves
(1/2) (g) (t²) = (4.9) (1.2²) = 7.06 meters .
c). & d).
1.2 seconds before it reaches maximum height, the projectile is located at
x = -14 m
y = -7.06 m
e). & f).
1.2 seconds after it reaches maximum height, the projectile is located at
x = +14 m
y = -7.06 m .
I hope you recognize that 6 answers, plus a little bit of explanation,
all for 5 points, ain't too shabby. You made out well.
5 N to the right
There’s a force of 6 N pulling to the right (the positive direction) and 1 N pulling left (the negative direction).
Net force= 6 N - 1 N = 5 N to the right
Answer:
Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-
Substituting the given values in above equation, we get -
Now we will determine the radius of curvature for the curve shown in the attached image
Differentiating on both the sides, we get -
![\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20-3.2%20%2810%5E-3%29%20X%5C%5C%5Cfrac%7Bd%5E2y%7D%7Bd%5E2x%7D%20%3D%20%20-3.2%20%2810%5E-3%29%5C%5CCurve%20%3D%20%5Cfrac%7B%5B1%2B%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%20%20%7D%7B%5Cfrac%7Bd%5E2y%7D%7Bd%5E2x%7D%7D%20%5C%5CCurve%20%3D%20312.5)
meter
Acceleration on curved path
Final acceleration
