To solve this problem, let’s analyze the motion of the ball step by step:

Given:

1. Initial height $h_0 = 5 \, \text{m}$
2. Initial velocity $u = 18 \, \text{m/s}$
3. Angle of projection $\theta = 40^\circ$
4. Time $t = 2 \, \text{s}$
5. Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$

We are tasked with finding the height of the ball above the ground after $2 \, \text{s}$.

Step 1: Resolve the initial velocity into components.

The velocity components are:

• Vertical component: $u_y = u \sin \theta$
• Horizontal component (not needed for height): $u_x = u \cos \theta$

Substitute $u = 18 \, \text{m/s}$ and $\theta = 40^\circ$: $u_y = 18 \sin 40^\circ \quad \text{and} \quad u_x = 18 \cos 40^\circ$ Using trigonometric values:

• $\sin 40^\circ \approx 0.6428$
• $\cos 40^\circ \approx 0.7660$

Thus: $u_y = 18 \times 0.6428 = 11.57 \, \text{m/s}$ (The horizontal velocity $u_x$ is irrelevant for this question.)

Step 2: Use the kinematic equation for vertical motion.

The vertical displacement $y$ after time $t$ is given by: $y = h_0 + u_y t - \frac{1}{2} g t^2$ Where:

• $h_0 = 5 \, \text{m}$ (initial height)
• $u_y = 11.57 \, \text{m/s}$
• $g = 9.8 \, \text{m/s}^2$
• $t = 2 \, \text{s}$

Substitute the values into the equation: $y = 5 + (11.57)(2) - \frac{1}{2} (9.8)(2^2)$

Simplify step by step:

1. $u_y t = 11.57 \times 2 = 23.14 \, \text{m}$
2. $\frac{1}{2} g t^2 = \frac{1}{2} (9.8)(4) = 19.6 \, \text{m}$

Thus: $y = 5 + 23.14 - 19.6$ $y = 8.54 \, \text{m}$

Final Answer:

The height of the ball above the ground after $2 \, \text{s}$ is 8.54 meters.

Would you like further clarification or details?

Here are 5 related questions for further practice:

1. What is the horizontal distance traveled by the ball after 2 seconds?
2. At what time will the ball reach its maximum height?
3. What is the maximum height reached by the ball?
4. What is the total time of flight of the ball before it hits the ground?
5. How far from the projection point will the ball land?

Tip: Always separate motion into horizontal and vertical components when solving projectile motion problems.