Question

a ball is thrawn up with a speed of 0.5m/s .
1.how high will it go before it begins to fall?
2.how long will it take to reach that height
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Answer 1

Answer:

Let H be the maximum height of the ball .

When , the ball is reached one half of the maximum height , then

          u=10m/s,v=0m/s,g=10m/s  

2

 and h=H/2 (remaining height) ,

by using ,    

             v  

2

=u  

2

−2gh  

             0  

2

=10  

2

−2×10×(H/2)

or           H=100/10=10m

Explanation:

I hope it will help you

pls mark brainliest.

Answer 2

Explanation:

1) when we throw a ball upwards ,there will be no 'v' ..so,

v = 0m/s

u = 0.5m/s

g = -10m/s²

v² - u² = 2as

0-u² = 2as

-(0.5)² = -2(10)(s)

-(0.25) = -20(s) ( here negative represents that the body is moving against gravity )

s = 0.0125m ..

2) v= u +at

-u = at

-(0.5) = -10 t

t = 0.05sec..