Science, asked by priya407527, 11 months ago

A ball is thrown up with a velocity of 15m/s. How high will it go before it begins to fall (g=9.8 m/s). ​

Answers

Answered by lBrainlykingl
2

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Here, u= 15m/s, v=0,g= 9.8 m/s^2,h=?

From v^2-u^2=2gh

h= v^2-u^2/2g

= 0–15^2/2*(-9.8)

=11.48m

Answered by TheNightHowler
3

Answer :-

The ball will reach upto the height of 11.46 metres.

Explanation :-

Given :

  • u = 15 m/s
  • v = 0
  • g = -9.8 m/ {s}^{2}

Now,

 {v}^{2}  =  {u}^{2}  + 2gs \\  =  > s =   \frac{ {v}^{2}  -  {u}^{2} }{2g}  \\  =  > s =  \frac{0 -  {(15)}^{2} }{2 \times ( - 9.8)}  \\  =  > s =   \frac{ - 225}{ - 19.6}  \\  =  > s =  \frac{225}{19.6}  \\  =  > s = 11.48 \: m

° The ball will reach upto the height of 11.48 metres.

#answerwithquality #BAL

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