A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 78 m horizontally
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The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.
Answer:
Explanation:
As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.
Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.
Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.
Now, from second law of motion,
So, acceleration = 10 N /5 kg = 2 m/s².
Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.
v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.
So Final velocity v = 0+(2×10)=20 m/s.
And the final position can be determined using the second equation of motion.
s = ut+1/2at²
Final position = (0×10)+(0.5×2×10×10)= 100 m.
So the final position is 100 m.
Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.