Question

a ball is thrown vertically upward from the top of tower of the height60 m with a speed of 20 m/s . calculate tge maximum height attained by the ball​

Answer 1

Answer:

t=5s

t=5sTime to reach maximum height can be obtained from v=u+at

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)t

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2s

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2)

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20m

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 m

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ )

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2 t

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2 t ′

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2 t ′ =3s

t=5sTime to reach maximum height can be obtained from v=u+at0=20+(−10)tt=2ss=ut+0.5at 2 =20(2)+0.5(−10)(2) 2 =20mThus, total distance for maximum height is 45 ms=ut+0.5at 2 45=0+0.5(10)(t ′ ) 2 t ′ =3sTotal time= 3+2= 5s

Step-by-step explanation:

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