Question

A body is thrown with a velocity \( (5\hat{i} + 6\hat{j}) \, \text{m/s} \). Its maximum height is (given that acceleration due to gravity = \( 10 \, \text{m/s}^2 \)):

• A. \(12.5 \, {m}\)
• B. \(1.8 \, {m}\)
• C. \(1.25 \, {m}\)
• D. \(0.9 \, {m}\)

Hint:

In projectile motion, only the vertical component of the initial velocity affects the maximum height reached. Remember, gravity only acts in the vertical direction and affects the vertical motion.

Answer 1

The Correct Option is B

To find the maximum height, we'll consider the vertical motion of the body.

1. Vertical component of velocity: The initial vertical velocity (\( u_y \)) is given by the \( j \)-component of the velocity vector:

\[ u_y = 6 \text{ m/s} \]

2. Kinematic equation: We can use the following kinematic equation to find the maximum height (\( H \)):

\[ v_y^2 = u_y^2 + 2as \]

where:

• \( v_y = 0 \) m/s (final vertical velocity at maximum height)
• \( u_y = 6 \) m/s (initial vertical velocity)
• \( a = -10 \) m/s\(^2\) (acceleration due to gravity, negative since it acts downwards)
• \( s = H \) (displacement, which is the maximum height)

3. Substitute and solve for \( H \):

\[ 0^2 = 6^2 + 2 \times (-10) \times H \] \[ 0 = 36 - 20H \] \[ 20H = 36 \] \[ H = \frac{36}{20} = 1.8 \text{ m} \]

Therefore, the maximum height reached by the body is 1.8 m.