From the edge of the rooftop of a building, a boy throws a stone at an angle 24.0° above the horizontal. The stone hits the ground 4.40 s later, 104 m away from the base of the building. (Ignore air resistance.) Find the initial velocity of the stone. What's the magnitude and direction? Find the initial height h from which the stone was thrown. Find the maximum height H reached by the stone.
To solve this problem, we can break it down into several steps:
Step 1: Analyze the given information.
- Angle of projection: 24.0° above the horizontal
- Time of flight: 4.40 s
- Horizontal displacement: 104 m
Step 2: Find the initial velocity of the stone.
We can use the equation of motion for horizontal displacement:
displacement = velocity * time
Since the horizontal displacement is 104 m and the time is 4.40 s, we can solve for the initial velocity:
104 m = velocity * 4.40 s
velocity (initial velocity) = 104 m / 4.40 s
Step 3: Calculate the initial velocity magnitude and direction.
To find the magnitude of the velocity, we can use the Pythagorean theorem:
magnitude = √(velocity_x^2 + velocity_y^2)
Since the stone is thrown at an angle above the horizontal, we need to find the horizontal and vertical components of the velocity.
velocity_x = velocity * cos(angle)
velocity_y = velocity * sin(angle)
Substituting these values into the equation, we get:
magnitude = √( (velocity * cos(angle))^2 + (velocity * sin(angle))^2 )
Step 4: Find the initial height h from which the stone was thrown.
To find the initial height, we can use the equation of motion for vertical displacement:
displacement = (initial velocity * sin(angle)) * time - (0.5 * gravity * time^2)
Since the stone reaches the ground, the vertical displacement is equal to zero. So we have:
0 = (initial velocity * sin(angle)) * time - (0.5 * gravity * time^2)
Simplifying the equation, we get:
initial velocity * sin(angle) = (0.5 * gravity * time^2) / time
initial velocity * sin(angle) = 0.5 * gravity * time
initial velocity = (0.5 * gravity * time) / sin(angle)
Step 5: Find the maximum height H reached by the stone.
To find the maximum height, we can use the equation of motion for vertical displacement:
displacement = (initial velocity * sin(angle)) * time - (0.5 * gravity * time^2)
At the maximum height, the vertical displacement is equal to the initial height h. So we have:
h = (initial velocity * sin(angle)) * time - (0.5 * gravity * time^2)
Substituting the value of the initial velocity from step 4, we get:
h = [(0.5 * gravity * time) / sin(angle)] * sin(angle) * time - (0.5 * gravity * time^2)
Simplifying the equation, we get:
h = 0.5 * gravity * time^2 - (0.5 * gravity * time^2)
h = 0
Therefore, the maximum height reached by the stone is 0.
Summarizing the results:
- The initial velocity of the stone is found to be 23.32 m/s.
- The magnitude of the initial velocity is 23.32 m/s.
- The direction of the initial velocity (angle) is 24.0° above the horizontal.
- The initial height from which the stone was thrown is 0 m.
- The maximum height reached by the stone is 0 m.
To find the initial velocity of the stone, we can use the equations of projectile motion.
First, we need to split the initial velocity into horizontal and vertical components.
Given:
Angle of projection: θ = 24.0°
Time of flight: t = 4.40 s
Horizontal displacement: x = 104 m
Step 1: Find the vertical component of the initial velocity (Vy).
The formula for the vertical component of the initial velocity (Vy) is given by:
Vy = V * sinθ
where V is the initial velocity.
To find Vy, we need to rearrange the equation as follows:
V = Vy / sinθ
Step 2: Find the horizontal component of the initial velocity (Vx).
The formula for the horizontal component of the initial velocity (Vx) is given by:
Vx = V * cosθ
where V is the initial velocity.
Step 3: Find the initial velocity (V).
We can use the horizontal displacement (x) and the time of flight (t) to find the initial velocity (V).
Vx = x / t
Now, we substitute the given values into the equations:
V = x / t = 104 / 4.40 = 23.64 m/s
Step 4: Calculate Vy.
Vy = V * sinθ = 23.64 * sin(24.0°) = 9.79 m/s
Step 5: Calculate Vx.
Vx = V * cosθ = 23.64 * cos(24.0°) = 21.43 m/s
Therefore, the initial velocity of the stone is 23.64 m/s, with a vertical component of 9.79 m/s and a horizontal component of 21.43 m/s.
To find the magnitude and direction of the initial velocity, we can use the Pythagorean theorem:
|V| = √(Vx^2 + Vy^2) = √(21.43^2 + 9.79^2) = 23.92 m/s
The direction of the initial velocity can be found using the inverse tangent function:
θ = tan^(-1)(Vy / Vx) = tan^(-1)(9.79 / 21.43) ≈ 25.7°
Therefore, the magnitude of the initial velocity is approximately 23.92 m/s, and the direction is approximately 25.7° above the horizontal.
Now let's find the initial height (h) from which the stone was thrown:
Using the formula for vertical displacement under constant acceleration:
h = Vy * t + 0.5 * g * t^2
where g is the acceleration due to gravity.
Since the stone was thrown horizontally, Vy = 0 initially, and the equation simplifies to:
h = 0.5 * g * t^2
Substituting the known values:
h = 0.5 * 9.8 m/s^2 * (4.40 s)^2
h ≈ 48.46 m
Therefore, the initial height from which the stone was thrown is approximately 48.46 m.
Finally, let's find the maximum height (H) reached by the stone:
Using the formula for vertical displacement under constant acceleration:
H = Vy^2 / (2 * g)
Substituting the known values:
H = (9.79 m/s)^2 / (2 * 9.8 m/s^2)
H ≈ 4.90 m
Therefore, the maximum height reached by the stone is approximately 4.90 m.