Question
a ball is projected at an angel of 45 degree to the horizontal.if the horizontal range is 40m,the maximum height attained by the ball is-
a)5m
b)8m
c)10m
d012m
a)5m
b)8m
c)10m
d012m
Answers
Elena
L=vₒ²•sin2α/g= vₒ²•2sinα•cosα/g
h= vₒ²•sin²α/2g,
h/L = vₒ²•sin²α•g / vₒ²•2sinα•cosα• 2g=
=tan α/4=tan45/4=1/4=0.25
h=0.25L=0.25•40=10 m
Answ. c) – 10 m
h= vₒ²•sin²α/2g,
h/L = vₒ²•sin²α•g / vₒ²•2sinα•cosα• 2g=
=tan α/4=tan45/4=1/4=0.25
h=0.25L=0.25•40=10 m
Answ. c) – 10 m