Question

A projectile is launched at an angle ‘α’ with the horizontal with a velocity \(20\) \(ms^{–1}\). After \(1 0\) \(s\), its inclination with horizontal is ‘β’. The value of tanβ will be (\(g\) = \(10\) \(ms^{–2}\)).

• A. tanα + 5secα
• B. tanα– 5secα
• C. 2tanα– 5secα
• D. 2tanα + 5secα

Answer 1

The Correct Option is B

\(V_y = 20 \times sin α-10 \times 10\)

\(V_x = 20 \;cos α\)

∴ \(tan β = \frac{V_y}{V_x} = \frac{20\; sin α -100}{20 \;cos α}\)

= \(tan\;α – 5sec\;α\)

Therefore, the correct option is (B): \(tan\;α – 5sec\;α\)