Question

a man throws a ball with a speed of 25 metre per second in upwards direction it falls back after 6 second find height achieved by the ball if a=10 m/s​

Answer 1

initial velocity (u)=21m/s

acceleration=-g=-10m/s

we know that,

${v}^{2} = {u}^{2} - 2gh$

At the max height v=0

$2gh = {u}^{2}$

$h = \frac{ {u}^{2} }{2g}$

$h = \frac{625}{20}$

$h = 31.25m$

I hope this helps ( ╹▽╹ )

Answer 2

✬ Maximum Height = 31.25 m ✬

Explanation:

Given:

• Initial Velocity (u) of ball = 25 m/s
• Final Velocity (v) of ball = 0 { at maximum height }
• Acceleration (a) = –10 m/s { Because ball is going upward }

To Find:

• How much height (s) is attained by the ball ?

Formula to be used:

• 2as = v² – u²

Solution: Putting the values on Formula, we get

$\implies{\rm }$ 2as = v² – u²

$\implies{\rm }$ 2 x (–10) x s = 0² – 25²

$\implies{\rm }$ –20s = – 625

$\implies{\rm }$ s = –625/–20

$\implies{\rm }$ s = 31.25 m

Hence, The Maximum height attained by the ball is 31.25 m.