Math Problem Statement

A projectile motion problem and Newton's second law applications.

Solution

Let's break down and solve the questions based on the image you've provided.

Questions 4–8:

A projectile motion problem with a ball being kicked and forming a parabolic trajectory. Initial velocity is v0=10m/sv_0 = 10 \, \text{m/s} at an angle of 3030^\circ to the horizontal.

4. Calculate the initial velocity components along the x-axis and y-axis:

The initial velocity components can be found using trigonometry:

  • v0x=v0cos(30)v_{0x} = v_0 \cdot \cos(30^\circ)
  • v0y=v0sin(30)v_{0y} = v_0 \cdot \sin(30^\circ)

Substituting v0=10m/sv_0 = 10 \, \text{m/s}:

v0x=10cos(30)v_{0x} = 10 \cdot \cos(30^\circ) v0y=10sin(30)v_{0y} = 10 \cdot \sin(30^\circ)

5. Find the velocity at point A:

At point A (which is at the peak of the trajectory), the vertical velocity is zero, but the horizontal velocity remains constant.

6. Determine the velocity at t=0.5secondst = 0.5 \, \text{seconds}:

The horizontal velocity stays the same, while the vertical velocity changes due to gravity (g=9.8m/s2g = 9.8 \, \text{m/s}^2):

vy=v0ygtv_{y} = v_{0y} - g \cdot t Substitute v0yv_{0y} from above and t=0.5secondst = 0.5 \, \text{seconds} to find the vertical velocity.

7. Calculate the maximum height reached:

At maximum height, the vertical velocity is zero. Using the kinematic equation:

vy2=v0y22ghmaxv_{y}^2 = v_{0y}^2 - 2g \cdot h_{\text{max}} Solve for hmaxh_{\text{max}}.

8. Calculate the range (horizontal distance) of the ball:

The total time of flight can be found by doubling the time it takes to reach the peak, and then the range is:

Range=v0xtotal time\text{Range} = v_{0x} \cdot \text{total time}


Question 9:

A block being pulled by three forces is shown, with F1=40NF_1 = 40 \, \text{N}, F2=26NF_2 = 26 \, \text{N}, and F3=14NF_3 = 14 \, \text{N}. We need to determine if the block is moving or stationary.

The total horizontal force is:

Ftotal=F1F2F3F_{\text{total}} = F_1 - F_2 - F_3 Calculate if FtotalF_{\text{total}} is greater than zero to see if the block moves.

Question 10:

A 10 kg block is being pulled by two forces: 5 N and 10 N. The net force is the difference between these forces, and the acceleration can be calculated using Newton's second law:

Fnet=F2F1=10N5N=5NF_{\text{net}} = F_2 - F_1 = 10 \, \text{N} - 5 \, \text{N} = 5 \, \text{N} a=Fnetm=510=0.5m/s2a = \frac{F_{\text{net}}}{m} = \frac{5}{10} = 0.5 \, \text{m/s}^2


Would you like me to calculate each value step by step, or provide details on a specific part?

Here are some related questions to consider:

  1. How does the angle of projection affect the range of a projectile?
  2. How would you find the time of flight for a projectile launched at an angle?
  3. What is the role of gravity in determining the vertical velocity of a projectile?
  4. How can you use Newton's second law to find the acceleration of an object with multiple forces acting on it?
  5. What happens if the net force on an object is zero? Does it always remain at rest?

Tip: In projectile motion, the horizontal and vertical motions are independent, with gravity only affecting the vertical component.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Newton's Laws of Motion

Formulas

v_x = v_0 * cos(θ)
v_y = v_0 * sin(θ) - g * t
Range = v_x * time
v_y^2 = v_0y^2 - 2gh
F_net = F_2 - F_1
a = F_net / m

Theorems

Newton's Second Law
Projectile motion formulas

Suitable Grade Level

Grades 10-12