Question

from the top a tower of height 20m a ball is thrown horizontally with a speed of 20m/s. If g=10m/s2 , the dot product of acceleration and velocity at the time when the ball reaches the ground
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Answer 1

Answer:

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Explanation:

Answer 2

Answer:

The dot product of acceleration and velocity at the time when the ball reaches the ground is 200 m²/s³.

Explanation:

Let the final velocity of the ball be (v) = $v_xi + v_yj$

Given,

The initial velocity of the ball (u) = 20 m/s

Height of the tower (h) = 20 m

acceleration of the ball before the ball hits the ground (a) = g j

To find,

$\bar{a}.\bar{v}$

Calculation,

We know that vertical velocity of the ball before it hits the ground is given by:

$v_y = 2gh$

$\implies v_y^2 = 2(10 m/s^2)(20 m)\\\implies v_y = \sqrt{400 }m/s$

      $v_y = 20 m/s$

Therefore, the vertical velocity before it hits the ground is 20 m/s.

Now, considering:

$\bar{a}.\bar{v} = (g j).(v_x i + v_yj)\\\implies \bar{a}.\bar{v} = g (v_y)\\\implies \bar{a}.\bar{v} = 10 m/s^2(20 m/s)\\\implies \bar{a}.\bar{v} = 200 m^2/s^3$

Therefore, the dot product of acceleration and velocity at the time when the ball reaches the ground is 200 m²/s³.

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