Question

please answer properly with explanation in detail and i will mark you the Brainliest if tge answer is correct​

Answer 1

Explanation:

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Given: time=6s,g=10m/s

Given: time=6s,g=10m/s 2

Given: time=6s,g=10m/s 2 ;

Given: time=6s,g=10m/s 2 ;To calculate the greatest height reached by the ball.

Given: time=6s,g=10m/s 2 ;To calculate the greatest height reached by the ball.Assume ‘h’ to be the greatest height reached. Ascent

Given: time=6s,g=10m/s 2 ;To calculate the greatest height reached by the ball.Assume ‘h’ to be the greatest height reached. AscentTime for the rise of the ball = 2/6 =3s

6 =3s Initial velocity for the rise is zero

6 =3s Initial velocity for the rise is zero As per the equation of motion;

$h = ut + \frac{1}{2} {gt}^{2}$

$= 0(3) + \frac{1}{2} (10)( {3)}^{2}$

$0 + 45 = > 45m$

Answer 2

Answer:

We will use second equation of motion

s=ut+1/2at^2

Explanation:

s=?

u=0

a=10m/s^2

t=3s. ( because time taken only for going is 3 second and for coming down it takes 6 seconds )

Now

s=ut+1/2at^2

s=0×3+1/2×10×3×3

s=0+45

s=45m

2.to find initial velocity we will use third equation of motion

v^2=u^2+2as

v=?

u=0

a=10m/s^2

s=45m

v^2=u^2+2as

v^2=0+2×10×45

v^2=900

v=30m (as we know that 900 is square root of 30)