Question

A ball is thrown up with a speed of 1m/s
I) how high will be go before it begins to fall ?
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Answer 1

Answer:

Explanation:

Final speed of ball ,v=0 (The ball stops ) Acceleration due to gravity g=−9.8m/s2 (Retardation )

And Height h=? (to be calculated )

Now ,Putting al these values in the formula

v2=u2+2gh

we get : (0)2=(15)2+2×(−9.8)×h

0=225-19.6 h

19.6 h =225

h=22519.6

h=11.4 m

Thus the ball will go the a maximum height of 11.4 meters before of begins to fall.