Answer:
Teh velocity of ball with respect to the ground is 29.2 m/s.
Explanation:
Velocity of car, vc = 25 due m/s east
velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis
Write the velocities in the vector form
The velocity of ball with respect to ground is
To solve this problem we will apply the concepts of energy conservation and Newton's second law that defines force as the product of the object's mass with its acceleration. Additionally we will apply concepts related to the kinematics equations of linear motion.
For conservation of energy we have that work is equal to kinetic energy therefore,
Here,
F = Force
d = Displacement
m = Mass
v= Velocity
At the same time we have the relation of
Therefore the value of the force can be interpreted as the rate of increase in energy per unit of distance, which makes it equivalent to
Applying Newton's Second Law
In 4 seconds final velocity of the object becomes
Then the work done is equal to,
Then the displacement is,
Therefore the distance moved is 16m
1) The landing spot of the projectile is given by
2) The common variable is the time
Explanation:
1)
The motion of a projectile consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
The landing spot can be determined in the following way:
- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation
where
h is the vertical displacement of the projectile, which corresponds to the height from which the projectile has been fired, above the ground
u = 0 is the initial vertical velocity
is the acceleration of gravity
t is the time of flight
Solving for t,
- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity . Therefore, the horizontal distance covered is given by
And substituting the time of flight,
2)
Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by
where is the horizontal velocity and t is the time.
The vertical motion is accelerated, so the vertical component of the displacement is given by
where g is the acceleration of gravity and t is the time.
Therefore, from the two equations we see that the common variable is t, the time.
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