A projectile is thrown at an angle of 30° with the horizontal so that its range is 200 m. Another projectile is thrown at an angle of 60° with the verticle . calculate it's range
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assuming the 30 degree angle is with horizontal.
assuming the initial velocity is u ( not equal to 0) as you said fired.
assuming the the height to be covered is h.
Now,
initial vel in horizontal direction ucos30 = [(3^1/2)/2 ]*u
initial vel in vertical direction usin30 = u/2.
distance to be covered in vertical = h
so by newton's laws time taken to reach h :
h = [(3^1/2)/2 ]*u*t + 0.5*t^2.
it is a quadratic equation in t , solve it yourself when you have the values of h and u.
vertical velocity after that time t = u + g*t (g is almost equal to 9.8)
tan(new angle) = vertical velocity/horizontal velocity
horizontal velocity will be unchanged.
so angle will be less then 30 degree with horizontal as g will increase vertical velocity.
If you had given me some values i'd have given you the exact answer but you presenly gave incomplete data.
assuming the initial velocity is u ( not equal to 0) as you said fired.
assuming the the height to be covered is h.
Now,
initial vel in horizontal direction ucos30 = [(3^1/2)/2 ]*u
initial vel in vertical direction usin30 = u/2.
distance to be covered in vertical = h
so by newton's laws time taken to reach h :
h = [(3^1/2)/2 ]*u*t + 0.5*t^2.
it is a quadratic equation in t , solve it yourself when you have the values of h and u.
vertical velocity after that time t = u + g*t (g is almost equal to 9.8)
tan(new angle) = vertical velocity/horizontal velocity
horizontal velocity will be unchanged.
so angle will be less then 30 degree with horizontal as g will increase vertical velocity.
If you had given me some values i'd have given you the exact answer but you presenly gave incomplete data.
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