A ball thrown by a boy in the street is caught 5 s later by another boy on the balcony of the house 10 m away and 3 m above the street level. What is the initial speed of the ball? What is the angle above the horizontal at which it was thrown?
If it takes 5 seconds to cover that distance you will have to throw it just about straight up so it will stay up five seconds !
u = constant horizontal speed = 10m/5s
= 2 meters/second
in 5 seconds it goes up 3 meters
3 = Vi t - 4.9 t^2
3 = Vi(5) = 4.9(25)
3 = 5 Vi - 122.5
Vi = 25.1 m/s
so
speed = sqrt (25.1^2 + 2^2)
= 25.2 m/s
tan T = Vi/u = 25.2/2
T = 85.5 degrees
Why did the ball bring a map to the balcony? Because it wanted to find its way to the house!
Now, let's get to business. To find the initial speed of the ball, we can use the formula for constant acceleration. In this case, the acceleration is just due to gravity, which is approximately 9.8 m/s².
First, we'll find the time it takes for the ball to reach the balcony. Since the ball was caught 5 seconds later, the time is t = 5 s.
Next, we'll find the horizontal distance the ball traveled. The horizontal distance can be calculated using the formula: x = v₀x * t, where v₀x is the initial horizontal velocity.
Given that the horizontal distance is 10 m and the time is 5 s, we can rearrange the formula to solve for v₀x: v₀x = x / t = 10 m / 5 s = 2 m/s.
Now, to find the initial vertical velocity, we'll use the formula: y = v₀y * t + (1/2) * a * t², where y is the vertical distance (3 m), v₀y is the initial vertical velocity, and a is the acceleration due to gravity (-9.8 m/s²).
Plugging in the values, we get: 3 m = v₀y * 5 s + (1/2) * (-9.8 m/s²) * (5 s)².
Simplifying the equation, we get: 3 m = v₀y * 5 s - 122.5 m.
Rearranging the equation, we find: v₀y * 5 s = 125.5 m.
Dividing both sides by 5, we get: v₀y = 25.1 m/s.
Now that we have the horizontal and vertical velocities, we can find the initial speed and the angle at which the ball was thrown.
The initial speed, v₀, can be calculated using the Pythagorean theorem: v₀ = sqrt(v₀x² + v₀y²).
Substituting the values, we get: v₀ = sqrt((2 m/s)² + (25.1 m/s)²) ≈ 25.2 m/s.
To find the angle above the horizontal, we can use the inverse tangent function: θ = arctan(v₀y / v₀x).
Substituting the values, we get: θ = arctan(25.1 m/s / 2 m/s) ≈ 87.6°.
So, the initial speed of the ball is approximately 25.2 m/s, and it was thrown at an angle of around 87.6° above the horizontal. Keep in mind that these values are approximate due to rounding.
To solve this problem, we can use the equations of motion to find the initial speed and angle.
Step 1: Calculate the horizontal component of the initial velocity.
The horizontal displacement is given as 10 m, and the time of flight is given as 5 s.
Using the equation: horizontal displacement = initial horizontal velocity * time
10 m = initial horizontal velocity * 5 s
initial horizontal velocity = 10 m / 5 s
initial horizontal velocity = 2 m/s
Step 2: Calculate the vertical component of the initial velocity.
The vertical displacement is given as 3 m, and the time of flight is given as 5 s.
Using the equation: vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
3 m = initial vertical velocity * 5 s + (1/2) * (-9.8 m/s^2) * (5 s)^2
3 m = initial vertical velocity * 5 s - 122.5 m
initial vertical velocity * 5 s = 122.5 m + 3 m
initial vertical velocity * 5 s = 125.5 m
initial vertical velocity = 125.5 m / 5 s
initial vertical velocity = 25.1 m/s
Step 3: Calculate the initial speed.
The initial speed can be found using the Pythagorean theorem.
initial speed = square root of (initial horizontal velocity^2 + initial vertical velocity^2)
initial speed = square root of (2 m/s)^2 + (25.1 m/s)^2
initial speed = square root of (4 m^2/s^2 + 630.01 m^2/s^2)
initial speed = square root of 634.01 m^2/s^2
initial speed = 25.2 m/s
Step 4: Calculate the angle above the horizontal.
The angle above the horizontal can be found using the tangent function.
angle = arctan (initial vertical velocity / initial horizontal velocity)
angle = arctan (25.1 m/s / 2 m/s)
angle = arctan (12.55)
angle ≈ 84.4°
Therefore, the initial speed of the ball is approximately 25.2 m/s, and the angle above the horizontal at which it was thrown is approximately 84.4°.
To find the initial speed of the ball, we can use the equations of motion in the horizontal and vertical directions. Let's break down the problem into components.
1. Horizontal Motion:
The ball travels a horizontal distance of 10 m in 5 seconds. We can use the equation:
Horizontal Distance = Initial Horizontal Velocity * Time
Since the initial horizontal velocity remains constant throughout the motion, we can simplify the equation to:
10 m = Initial Horizontal Velocity * 5 s
Rearranging the equation, we get:
Initial Horizontal Velocity = 10 m / 5 s = 2 m/s
2. Vertical Motion:
The ball travels vertically 3 m above the street level. We can use the equation:
Vertical Distance = Initial Vertical Velocity * Time + (0.5 * Acceleration * Time^2)
The vertical velocity initially is 0 m/s as the ball starts from rest vertically. The acceleration due to gravity is -9.8 m/s^2 (considering downward as the negative direction). We can rearrange the above equation to solve for the initial vertical velocity.
3 m = 0 * 5 + (0.5 * -9.8 * (5^2))
Rearranging and solving for the initial vertical velocity, we get:
Initial Vertical Velocity = (3 m - 0.5 * -9.8 * (5^2)) / 5
Initial Vertical Velocity = 24.5 m/s
Now, we can find the initial speed of the ball using the Pythagorean theorem, which states that the square of the hypotenuse (initial speed) is equal to the sum of the squares of the two sides (initial horizontal velocity and initial vertical velocity):
Initial Speed^2 = (Initial Horizontal Velocity)^2 + (Initial Vertical Velocity)^2
Substituting the known values:
Initial Speed^2 = (2 m/s)^2 + (24.5 m/s)^2
Initial Speed^2 = 4 m^2/s^2 + 600.25 m^2/s^2
Initial Speed^2 = 604.25 m^2/s^2
Taking the square root of both sides, we get:
Initial Speed = √(604.25 m^2/s^2)
Initial Speed ≈ 24.6 m/s (rounded to one decimal place)
Therefore, the initial speed of the ball is approximately 24.6 m/s.
To find the angle above the horizontal at which the ball was thrown, we can use trigonometry. The angle can be found using the equation:
tan(angle) = (Initial Vertical Velocity) / (Initial Horizontal Velocity)
Substituting the known values:
tan(angle) = 24.5 m/s / 2 m/s
tan(angle) ≈ 12.25
To find the angle, we take the inverse tangent (arctan) of both sides:
angle = arctan(12.25)
Using a scientific calculator or math software, we find:
angle ≈ 86.9 degrees (rounded to one decimal place)
Therefore, the angle above the horizontal at which the ball was thrown is approximately 86.9 degrees.