Question

The position of a projectile launched from the origin at t = 0 is given by r = (40hati + 50hatj)m at t = 2s. If the projectile was launched at an angle θ from the horizontal (take g = 10m s⁻2), then θ is

• A. \( \tan^{-1}\dfrac{2}{3} \)
• B. \( \tan^{-1}\dfrac{3}{2} \)
• C. \( \tan^{-1}\dfrac{7}{4} \)
• D. tan⁻1(4)/(5)

Hint:

Always resolve projectile motion into horizontal and vertical components.

Answer 1

The Correct Option is B

Step 1: Horizontal motion: 40 = ucosθ × 2 ⟹ ucosθ = 20
Step 2: Vertical motion: 50 = usinθ × 2 - (1)/(2)gt² 50 = 2usinθ - 20 ⟹ usinθ = 35
Step 3: Hence, tanθ = (35)/(20) = (7)/(4) ⟹ θ = tan⁻1(7)/(4)