Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
The answer to the question is 21 km towards east.
CALCULATION:
Let us consider two vectors 
As per the question, 
It is obvious that two vectors are parallel to each other. Hence, the angle between them is zero i.e 
We are asked to calculate the resultant .
As per parallelogram law of vector addition, the magnitude of resultant will be-
[∵ cos0 = 1]
The direction of the resultant will be also towards east as the two vectors are parallel to each other.
Hence, the vector addition of two vectors are 21 km towards east.
Argon because it is a noble gas
1. Answer;
Velocity decreases as it rises
Explanation;
When you project an object upward and release it at its initial velocity, it is moving in the opposite direction of the force of gravity. Thus the initial velocity is negative. The velocity of the object is also negative on the way up but positive on the way down.
2. Answer;
Velocity is 0 m/s at its highest point
Explanation;
The maximum height of the projectile is when the projectile reaches zero vertical velocity. From this point the vertical component of the velocity vector will point downwards.
The object slows down as it moves upward until it reaches a maximum height, at which time the velocity is zero. Then the velocity increases as the object falls toward the ground.
3. Answer;
-Acceleration is constant -9.8m/s^2
Explanation;
-An object that is thrown vertically upwards decelerates under the earth's gravity. Its speed decreases until it attains a maximum height, where the velocity is zero. Then it is accelerated uniformly downwards under gravity.
4. Answer;
-Acceleration is constant -9.8m/s^2
Explanation;
When a ball is thrown upwards the acceleration due to gravity remains constant throughout. It comes to rest for a moment at the highest point of motion just before returning back to earth.
-The acceleration due to gravity is constant on the object though out its flight. So the acceleration of the projectile is equal to the acceleration due to gravity, 9.81 meters/second/second, from just after its thrown, through its highest point, and until just before it hits the ground.
Answer:
0.5 m/s
Explanation:
From Newton's Third law of motion,
Momentum of the cannon = momentum of the flatcar.
mv = MV........................ Equation 1
Where m = mass of the cannon, v = velocity of the cannon, M = mass of the flatcar, V = Velocity of the flat car.
make V the subject of the equation
V = mv/M................... Equation 2
Given: m = 10 kg, v = 50 m/s, M = 1000 kg
Substitute into equation 2
V = 10×50/1000
V = 500/1000
V = 0.5 m/s
Hence the speed of the flatcar = 0.5 m/s
