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
2
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
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/
Now,
•°• The ball will reach upto the height of 11.48 metres.
#answerwithquality #BAL
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