Question

pls ans ques no. 5 its urgent

Answer 1

Total Mechanical Energy (TME) = KE +PE;
_________________________
initially ball is at ground (PE =0) and having velocity 20m/s (
KE = ½×m×v²
KE = ½× (15/1000)×(20)²
KE = 3J. )

Hence TME(initial) = 0+3 = 3J;
__________________________
we want a point at a height (say h) where
PE = KE .
As TME remains conserved ;
3= PE+KE
3 = 2×PE. (as PE = KE )
=> PE = 3/2 .
This is the PE (Potential Energy) at that point which is at height 'h'.
__________________________
We know that ;
PE = mgh ;
3/2= (15/1000)×10×h;
=> h= 10meters.
___________________________
Hope it helps!!