The acceleration of the rock is the gravitational acceleration due to earth which is around 9.8 m/s^2. As the rock travels upwards it is working against the constant force of gravity and when it reaches the top it is the force of gravity that will cause a cause in the direction of the rocks velocity and cause it to accelerate towards earth. Using the equation a = F/m, the acceleration due to gravity exerts a force per unit mass on an object causing the object to accelerate, anything above earth surface will therefore have an external gravitational force action on it (in terms of projectile motion). So at the top of the rocks trajectory, the acceleration of the rock will be g (9.8) as a constant force is gravity is exerted on the rock throughout its path.
Answer:
13 u2 = 0 – (2 × –2 m/s2 × 6 m) = 24, u = 4.9 m
Explanation:
Answer:
physical change uwu owo hope this helps you uwu silly goose
First compute the resultant force F:
Then use Newton's second law to determine the acceleration vector 
for the particle:
Let 
and 
denote the particle's position and velocity vectors, respectively.
(a) Use the fundamental theorem of calculus. The particle starts at rest, so 
. Then the particle's velocity vector at <em>t</em> = 10.4 s is
If you don't know calculus, then just use the formula,
So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component
(b) Compute the angle 
for 
:
so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.
(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):
Then using the fundamental theorem of calculus again, we have
where 
is the particle's initial position. So we get
So over the first 10.4 s, the particle is displaced by the vector
or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).
(d) See part (c).
