Question

a ball is thrown up with the speed of 10m/s . what will be it's final velocity and why?how long will it go before it begins to fall how long will it take to reach that height​

Answer 1

Hey mate here's your answer in detail:-

When the velocity is zero, we can assume in most cases that the object is at its maximum height. We also know that velocity is the first derivative of position, and acceleration is the second derivative of velocity:

x'(t) = v(t)

x''(t) = v'(t) = a(t)

Using what we know about the acceleration constant (I will use -10m/ss), we get the system of equations:

a(t) = -10

v(t) = 10

x(t) = M

Now, we can integrate, S -10 dt, and then we get -10t + C, and by calculus, this equals velocity. Now, by plugging in 10 for C, and knowing velocity must be zero to get the max height, we now have the equation, -10t + 10 = 0, where t is time. Solving it, we get 1 second to get to max height, answering the first question. Now, we integrate the velocity equation: S (-10t + 10) dt, and get x(t) = -5tt + 10t + C, where C is the initial height. Since you gave no height, I will just use C as 0.

The new equation is x(t) = -5tt + 10t. Since we figured time as one second, we plug it in, and get the answer, 5 meters, answering the other question. In a nutshell, integrate acceleration to velocity, plug initial velocity, set it equal to zero, get the time. Once we know time, we can do more integration to get the maximum height.

Answer 2

Explanation:

v2=u2+2as,v2=u2+2as, where v,u,av,u,a and ss are the final velocity, initial velocity, acceleration and displacement.

It is given that u=10u=10 m/s.

At the highest point, the velocity of the ball is zero 

⇒v=0.

⇒v=0.

The acceleration, a,a, is the acceleration due to gravity, which is −9.81−9.81 m/s2.2. It is negative because the direction of the acceleration due to gravity is opposite to the direction of motion.

⇒0=102−2×9.81s

⇒s=1002×9.81=5.1

⇒0=102−2×9.81s

⇒s=1002×9.81=5.1 m.

⇒ The ball goes up to a height of 5.1m.

Another method:

Initially the ball has only kinetic energy, which is 12mv2=12m(10)2=50m12mv2=12m(10)2=50m J.

Here vv is the initial velocity when the ball is just thrown up.

At the highest point, the ball has only potential energy, which is mgh=m(9.81)(h)=9.81mhmgh=m(9.81)(h)=9.81mh J.

By the principle of conservation of energy, 12mv2=mgh.12mv2=mgh.

⇒50m=9.81mh

⇒h=50m/9.81m=5.1

⇒ The ball goes up to a height of 5.1 m.