Math Problem Statement

You throw the ball from a wndow that is 0.9m above the ground at an angle of 12degree downwards from the horizontal plane. How far from the wall of the house does the ball hit the ground if its initital speed is 12.7 m/s Air resistance is not taken into account

Solution

In this problem, we're given the initial conditions of a projectile motion scenario:

  • The ball is thrown from a height of 0.9m0.9 \, \text{m}.
  • The initial speed of the ball is 12.7m/s12.7 \, \text{m/s}.
  • The angle of projection is 1212^\circ downwards from the horizontal.
  • We're ignoring air resistance.

We aim to determine how far from the wall (i.e., horizontal distance) the ball hits the ground.

Step 1: Break the Initial Velocity into Components

Since the ball is thrown at an angle of 1212^\circ below the horizontal, we can decompose the initial velocity v0=12.7m/sv_0 = 12.7 \, \text{m/s} into horizontal and vertical components:

  • Horizontal component: v0x=v0cos(12)v_{0x} = v_0 \cos(12^\circ)
  • Vertical component: v0y=v0sin(12)v_{0y} = v_0 \sin(12^\circ)

Step 2: Use the Kinematic Equation for Vertical Motion

We know the initial height h=0.9mh = 0.9 \, \text{m}, and the acceleration due to gravity g=9.8m/s2g = 9.8 \, \text{m/s}^2. The vertical displacement when the ball hits the ground is 0.9m-0.9 \, \text{m}, since it falls below the release point.

We can use the following kinematic equation for vertical motion:

y=v0yt+12gt2y = v_{0y} t + \frac{1}{2} g t^2

where:

  • y=0.9my = -0.9 \, \text{m} (the change in vertical height),
  • v0y=v0sin(12)v_{0y} = -v_0 \sin(12^\circ) (since the ball is projected downwards),
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2.

Step 3: Solve for Time of Flight

Solving this quadratic equation for tt, the time it takes for the ball to hit the ground, will allow us to find the horizontal distance.

Step 4: Calculate the Horizontal Distance

Once we know the time tt, we can compute the horizontal distance dd using the horizontal velocity component:

d=v0xt=v0cos(12)td = v_{0x} t = v_0 \cos(12^\circ) t

Now let's calculate.The ball will hit the ground approximately 9.64 meters away from the wall of the house.

Would you like more details or have any other questions?

Here are five related questions for further practice:

  1. How would the horizontal distance change if the initial speed were increased to 15 m/s?
  2. What happens to the time of flight if the ball is thrown from a higher window, say 1.5 meters above the ground?
  3. How does air resistance affect the trajectory and distance traveled?
  4. If the ball were thrown upwards at 12 degrees instead of downwards, how far would it travel horizontally before hitting the ground?
  5. How does the angle of projection affect both the horizontal distance and time of flight?

Tip: Always decompose projectile motion problems into horizontal and vertical components to simplify your calculations.

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Math Problem Analysis

Mathematical Concepts

Projectile motion
Kinematics
Trigonometry

Formulas

v_{0x} = v_0 * cos(θ)
v_{0y} = v_0 * sin(θ)
y = v_{0y} * t + (1/2) * g * t^2
d = v_{0x} * t

Theorems

Newton's laws of motion
Kinematic equations

Suitable Grade Level

High School (Grades 10-12)