Question

a stone dropped from the ground a height of 45m above the ground calculate the velocity of the ball just before it strike the ground neglect air resistance and the g=10m/so​

Answer 1

`30\; \rm m\cdot s^(-1)`.

Step-by-step explanation:

Under the assumptions, the acceleration of this ball would be constant during the descent. Thus, the SUVAT equations would apply.

Let
`x` denote the displacement of this stone during the fall. Let
`v_(0)` and
`v_(1)` denote the velocity of this ball at the beginning at the end of the fall, respectively. Let
`a` denote the acceleration of this ball. The SUVAT equation
`{v_(1)}^(2) - {v_(0)}^(2) = 2\, a\, x` would apply.

Rearrange the equation to find an expression for
`v_(1)`, the velocity of this ball at the end of the fall:

`\displaystyle {v_(1)}^(2) = 2\, a\, x + {v_(0)}^(2)`.

If the final velocity of the ball is in the same direction as the displacement (downwards), then
`v_(1) > 0`, such that
`\displaystyle v_(1) = \sqrt{2\, a\, x + {v_(0)}^(2)}`.

The acceleration of this ball during the fall should be equal to the gravitational field strength:
`a = g = 10\; \rm m\cdot s^(-2)`.

The displacement of the ball during the fall would be
`x = 45\; \rm m`.

The question states that the ball was "dropped" from the given height. Thus, the initial velocity of the ball would be
`0\; \rm m\cdot s^(-1)`. (That is,
`v_(0) = 0\; \rm m \cdot s^(-1)`.)

Substitute these values into the expression for
`v_(1)` and evaluate:

`\begin{aligned}v_(1) &= \sqrt{2\, a\, x + {v_(0)}^(2)} \\ &= \sqrt{2 * 10\; \rm m\cdot s^(-2) * 45\; \rm m + 0\; \rm m \cdot s^(-1)} \\ &= 30\; \rm m\cdot s^(-1)\end{aligned}`.

Therefore, the velocity of this ball right before landing would be
`v_(1) = 30\; \rm m\cdot s^(-1)`.