A projectile is thrown at an angle theta with the horizontal and an initial speed you show that the trajectory of the projectile is parabolic at what angle the range and maximum height becomes equal also show that range will be same for the complementary angles
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a= at tan-1(4) means 76° range and height both become equal.
nd 15+75=90
30+60= 90
45+45= 90 ok
so range will be same for these complementary angles like
R= v^2sin2@/g
= take v as 2
R= 2^2 sin 2*30/9.8
R= 4 sin 60 /9.8
sin 60= 0.866
R= 4* 0.866/9.8
= 0.353
now for 60
R= 2^2 sin 2*60/g
= 4*sin120/9.8
= sin 120 value is also 0.866
so 4*0.866/9.8
= 0.353
hope this helps☺
nd 15+75=90
30+60= 90
45+45= 90 ok
so range will be same for these complementary angles like
R= v^2sin2@/g
= take v as 2
R= 2^2 sin 2*30/9.8
R= 4 sin 60 /9.8
sin 60= 0.866
R= 4* 0.866/9.8
= 0.353
now for 60
R= 2^2 sin 2*60/g
= 4*sin120/9.8
= sin 120 value is also 0.866
so 4*0.866/9.8
= 0.353
hope this helps☺
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