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 (\(g\) = \(10\) \(ms^{–2}\)).
- tanα + 5secα
- tanα– 5secα
- 2tanα– 5secα
- 2tanα + 5secα
The Correct Option is B
Solution and Explanation
\(V_y = 20 \times sin α-10 \times 10\)
\(V_x = 20 \;cos α\)
∴ \(tan β = \frac{V_y}{V_x} = \frac{20\; sin α -100}{20 \;cos α}\)
= \(tan\;α – 5sec\;α\)
Therefore, the correct option is (B): \(tan\;α – 5sec\;α\)
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Concepts Used:
Speed and Velocity
The rate at which an object covers a certain distance is commonly known as speed.
The rate at which an object changes position in a certain direction is called velocity.
Difference Between Speed and Velocity:
Read More: Difference Between Speed and Velocity