Question

The position of a projectile launched from the origin at \(t=0\) is given by \[ \vec{r} = (40\hat{i}+50\hat{j})\,\text{m} \quad \text{at } t=2\,\text{s}. \] If the projectile was launched at an angle \(\theta\) from the horizontal (take \(g=10\,\text{m s}^{-2}\)), then \(\theta\) is:

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

Hint:

Use position components at a given time to back-calculate velocity components.

Answer 1

The Correct Option is B

Step 1: Horizontal motion: \[ x = u\cos\theta \, t \Rightarrow 40 = 2u\cos\theta \Rightarrow u\cos\theta = 20 \]
Step 2: Vertical motion: \[ y = u\sin\theta \, t - \frac{1}{2}gt^2 \Rightarrow 50 = 2u\sin\theta - 20 \] \[ \Rightarrow u\sin\theta = 35 \]
Step 3: Taking ratio: \[ \tan\theta = \frac{35}{20} = \frac{7}{4} \]