Question

A body is projected with a velocity of \(15\sqrt3\,\text{m s}^{-1}\) at an angle of \(60^\circ\) with the horizontal and another body is projected simultaneously from the same point in the same vertical plane with a velocity of \(40\,\text{m s}^{-1}\) at an angle of \(30^\circ\) with the horizontal. The time at which the velocity vectors of the two bodies will be in the same direction is:

• A. \(3.2\,\text{s}\)
• B. \(2.4\,\text{s}\)
• C. \(1.2\,\text{s}\)
• D. \(3.6\,\text{s}\)

Hint:

Whenever two velocity vectors are said to be parallel or in the same direction, equate their slopes: \[ \frac{v_y}{v_x} \] for both bodies.

Answer 1

The Correct Option is B

Step 1: What does same direction mean.
Two velocity vectors point the same way when they have the same slope, that is the same ratio of vertical part to horizontal part. So our goal is to make these two ratios equal and solve for time.

Step 2: Velocity parts of the first body.
It is thrown at $u_1 = 15\sqrt3$ at $60^\circ$. The horizontal part stays fixed, the vertical part drops with gravity. \[ v_{1x} = 15\sqrt3\cos60^\circ = \frac{15\sqrt3}{2} \] \[ v_{1y} = 15\sqrt3\sin60^\circ - 10t = \frac{45}{2} - 10t \]
Step 3: Velocity parts of the second body.
It is thrown at $u_2 = 40$ at $30^\circ$. \[ v_{2x} = 40\cos30^\circ = 20\sqrt3 \] \[ v_{2y} = 40\sin30^\circ - 10t = 20 - 10t \]
Step 4: Set the two slopes equal.
\[ \frac{\frac{45}{2} - 10t}{\frac{15\sqrt3}{2}} = \frac{20 - 10t}{20\sqrt3} \] To clear the half, write the top of the left side as $\dfrac{45 - 20t}{2}$ and combine.
Step 5: Cross multiply and simplify.
After clearing fractions the $\sqrt3$ cancels on both sides. \[ 20(45 - 20t) = 15(20 - 10t) \] \[ 900 - 400t = 300 - 150t \]
Step 6: Solve for time.
Bring the $t$ terms together and the numbers together. \[ 600 = 250t \] \[ t = 2.4\,\text{s} \] \[ \boxed{2.4\,\text{s}} \]