Question

12) A ball drop from tower to the ground in 0.5 sec. find (i) Velocity

y of ball when it strikes the ground. (ii) How high the tower from the ground. (a=10 m/s) solution .
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Answer 1

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Answer 2

Given:

Time elapsed by the ball to reach the ground (t) = 0.5 sec

Acceleration due to gravity (g or a) = 10 m/s

To Find:

• The Velocity of the ball when striking the ground
• Height of the tower

Solution:

For the Height of the tower,

To find the velocity, we have to find the height of the tower first.

So, the formula used will be,

                                       $s=ut+\frac{1}{2} at^{2}$

Here, s is the vertical displacement {i.e., Height of the tower

         u is the initial velocity,

         a is the acceleration,

 and, t is the time taken

By using the formula,

                         $s=(0)(0.5) + \frac{1}{2} (10)(0.5)^2$         {∵ ball is dropped from the rest

                         $s=0+\frac{1}{2}(2.5)$

                         $s=\frac{2.5}{2}\\s=1.25$

So, the Height of the tower is 1.25 m.

To find the velocity of the ball when it is striking the ground, we have to use the formula with the height of the tower,

                            $v^2=u^2+2as$

                            $v^2=(0)^2+2(10)(1.25)$      {∵ ball is dropped from the rest

                            $v^2=2(12.5)\\v^2=25\\v=\sqrt{25} \\v=5$    

So, the velocity when the ball hits the ground is 5 m/s