Question

Two bodies are thrown from the same point with the same velocity of $50\,m{{s}^{-1}}$. If their angles of projection are complementary to each other and the difference of maximum heights is 30 m, the maximum heights are $(g=10{m}/{{{s}^{2}}}\;)$
A. 50 m and 80 m
B. 47.5 m and 77.5 m
C. 30 m and 60 m
D. 25 m and 55 m

Answer 1

The formula that relates the parameters, such as height, the angle of projection and the initial velocity should be used to solve this problem. We will consider 2 equations of height and will substitute the values. The difference between these heights is given, so, we will compute the sum of heights. Finally using these equations, we will compute the height values.
Formula used:
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$

Complete answer:
From the given information we have the data as follows.
Two bodies are thrown from the same point with the same velocity of $50\,m{{s}^{-1}}$. Their angles of projection are complementary to each other and the difference of maximum heights is 30 m.
${{H}_{2}}-{{H}_{1}}=30\,m$
The complementary angles are the sine angle and the cosine angle.
Consider the formula for computing the height of a projection.
$H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$
Where u is the initial velocity, $\theta$is the angle of inclination and g is the acceleration due to gravity.
Now consider the expression for the heights.
${{H}_{1}}=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$ and ${{H}_{2}}=\dfrac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}$
Substitute the values in the above equations.
${{H}_{1}}=\dfrac{{{50}^{2}}{{\sin }^{2}}\theta }{2\times 10}$ and ${{H}_{2}}=\dfrac{{{50}^{2}}{{\cos }^{2}}\theta }{2\times 10}$
Add the above equations.

& {{H}_{2}}+{{H}_{1}}=\dfrac{{{50}^{2}}{{\sin }^{2}}\theta }{2\times 10}+\dfrac{{{50}^{2}}{{\cos }^{2}}\theta }{2\times 10} \\\ & \Rightarrow {{H}_{2}}+{{H}_{1}}=\dfrac{{{50}^{2}}}{2\times 10}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta ) \\\ & \therefore {{H}_{2}}+{{H}_{1}}=\dfrac{{{50}^{2}}}{2\times 10}=125\,m \\\ \end{aligned}$$ Consider the equations. $${{H}_{2}}+{{H}_{1}}=125\,$$…… (1) $${{H}_{2}}-{{H}_{1}}=30$$……. (2) Solve the equations (1) and (2). $$\therefore {{H}_{1}}=47.5\,cm,{{H}_{2}}=77.5\,cm$$ $$\therefore $$The values of the heights from which the bodies are thrown are 47.5 m and 77.5 m . **Thus, option (B) is correct.** **Note:** As the motion curve involved was a projection, thus, the height of the projection formula is used. The other complementary angles are: the tan angle and cot angle, the cosine angle and the secant angle.