Question

hey help me doing this quest9​

Answer 1

Q.9

Given that

Initial velocity of stone = 5 m/s

Acceleration = 10 m/s² in downward direction

Now as we know at Max height

Final velocity = 0

Then from motion equation

v² = u² + 2as

=> 0² = 5² + 2(-10)s

=> -25 = -20 s

or s = $\frac{-25}{-20}$

= $\frac{5}{4}$

= 1.25 m

hope it helps

Answer 2

Here is your answer of Question 9 :

Give.:

initial velocity of stone (u)= 5 m/s

acceleration (a) = -10 m/s² ( force of attraction is acting in downward direction )

We know that the final velocity of the stone at maximum height is (v) = 0 m/s

To Find:

Distance (s) = ?

Applying third equation of motion

v² = u² + 2as

Put the values

0² = 5² + 2(-10)*s

s = -25/-20

s = 5/4

s = 1.25 m/s