Question

Two objects are projected at an angle θ° and (90-θ)°, to the horizontal with the same speed. The ratio of their maximum vertical height is

• A. 1 : Tan θ
• B. 1 : 1
• C. tan2 θ : 1
• D. tan θ : 1

Answer 1

Answer: (C)

tan2 θ : 1

Answer 2

Let's consider the motion of the two objects separately.
For the object projected at an angle θ:
Initial vertical velocity (uy1) = u sin θ
Initial horizontal velocity (ux1) = u cos θ
For the object projected at an angle (90-θ):
Initial vertical velocity (uy2) = u sin (90-θ) = u cos θ
Initial horizontal velocity (ux2) = u sin (90-θ) = u sin θ
We can calculate the maximum vertical height reached by each object using the equation for vertical displacement:
h_max = (uy2) / (2g)
For the object projected at angle θ:
hmax1 = (uy12) / (2g) = (u sin θ)2 / (2g) = u2 * sin2 θ / (2g)
For the object projected at angle (90-θ):
hmax2 = (uy22) / (2g) = (ucosθ)2 / (2g) = u2 x cos2 θ / (2g)
To find the ratio of their maximum vertical heights, we can divide hmax1 by hmax2:
hmax1 / hmax2 = (u2 x sin2 θ / (2g)) / (u2 x cos2 θ / (2g))
Canceling out the common factors of u2 and 2g, we get:
hmax1 / hmax2 = sin2 θ / cos2 θ = tan2 θ
Therefore, the ratio of their maximum vertical heights is tan2 θ : 1 (option C).