# A ball is dropped from a height of 18 m. What is the speed of the ball when it touches the ground. Determine the speed using two different ways ? (I got the answer but what can be a second way)

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A ball is dropped from a height of 18 m. The speed of the ball when it touches the ground has to be determined.

An object with an initial velocity u m/s, accelerating at a m/s^2 has a velocity v after it travels a distance s where v^2 - u^2 = 2*a*s

When the ball is dropped from a height of 16 m, the acceleration due to the gravitational force is 9.8 m/s^2 and the distance traveled is 16 m. The initial velocity is 0. If the velocity of the ball when it lands is v, v^2 = 2*9.8*16 + 0^2

=> v = `sqrt(2*9.8*16)`

=> v `~~` 17.7 m/s

Another way of determining the velocity of the ball when it lands on the ground is by using the law of conservation of energy. At a height of 16 m, the ball is at rest and does not have any kinetic energy. If the mass of the ball is m, the potential energy of the ball is equal to m*g*h = m*9.8*16. When the ball hits the ground, if its speed is v, the kinetic energy is (1/2)*m*v^2 and potential energy is 0.

Equating the two m*9.8*16 = (1/2)*m*v^2

=> v = `sqrt(2*9.8*16)`

=> v `~~` 17.7 m/s

The result using both methods is the same. The speed of the ball when it strikes the ground is approximately 17.7 m/s

You can take two different approaches, although the final answer will be the same:

(1) **Classical Mechanics:**

v^2 = u^2 + 2as

where v = final velocity, u = initial velocity, a = acceleration, and s = distance covered

Here u = 0

So, v^2 = 2as

or v =sqrt(2as)

= sqrt(2*9.8*18) = sqrt(352.8) = **18.78 m/s**

(2) **Conservation of energy:**

When the ball is at a height of h, it has a potential energy = mgh

When the ball hits the ground with a velocity v, it has a kinetic energy

= 1/2 mv^2

Due to conservation o energy, these two should be equal.

So, 1/2 mv^2 = mgh

or 1/2 v^2 = gh

or v^2 = 2gh

or v = sqrt(2gh)

= sqrt(2*9.8*18) = sqrt(352.8) = **18.78 m/s**

Actually I only required the second part, i.e. the application of law of conservation of energy. It was quite evident that you had made a typo in deriving the speed on the ground, not an issue.

`1)`

` v^(2)=u^(2) + 2as` `v^(2)= 0 + 2*(9.8)*(18)`

`v = sqrt(2*(9.8)*(18))= 18.78ms^(-1)`

` 2)`

`mgh = (1)/(2) mv^(2)`

`gh = 1/2 v^(2)`

`v = sqrt(2gh)`

`v= sqrt(2*(9.8)*(18)) = 18.78ms^(-1)`

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One method:

Equation of motion:

(v^2) = (u^2)-2as

Here v is final velocity; u is initial velocity; a is acceleration and s is displacement or distance travelled.

Here,ball is dropping from a certain height.The acceleration is thus negative.The equation can be transformed into:

v^2 = (u^2)+2as

v = ?, u = 0, a = 9.81m/s^2,s = 18m

Thus,

v^2 = 0+2×9.81×18

v = sqrrt (2×9.81×18)

= 18.79m/s

Another method:

For the body of mass m,

Net potential energy = net kinetic energy

mgh2-mgh1 = ((1/2)×m×v^2)-((1/2)×m×u^2)

h1 = 0,h2 = 18m; v = ?,u = 0

Cancelling m on both sides and substituting the values in the equation, we get,

9.81×18 = (1/2)×v^2

v = sqrrt(2×9.81×18)

18.79m/s