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

Answer: (B)

tanα– 5secα

Answer 2

$V_y = 20 \times sin α-10 \times 10$

$V_x = 20 \;cos α$

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

= $tan\;α – 5sec\;α$

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