A car weighs 2,000 kg. It moves along a road by applying a force on the road with a parallel component of 560 N. There are two p
1 answer:
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Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity () = 0 m/s
Final velocity () = ?
Height = 2 m
and, = 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions
235.44 = + 78.48
= 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.
<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.