A ball thrown horizontally at 13.0m/s from the roof of a building lands 21.4m from the base of the building. How tall is the building?

first do the horizontal problem with constant horizontal speed of 13 for 21.4 m to get time in air

distance = speed * time
t = 21.4 /13.0

now we know it fell for time t

h = (1/2) g t^2
h = (1/2) (9.81) (21.4/13.0)^2 meters

To find the height of the building, we can use the formula for the horizontal range of a projectile:

Range = Initial velocity * Time

In this case, the initial velocity in the horizontal direction is given as 13.0 m/s, and the range is given as 21.4 m. The time is not given directly, but we can find it using the formula for vertical motion.

The formula for the vertical displacement of a projectile is:

Vertical displacement = Initial vertical velocity * Time + (1/2) * Acceleration * Time^2

In this case, the initial vertical velocity is 0 m/s since the ball is thrown horizontally, and the acceleration is due to gravity, which is approximately -9.8 m/s^2.

Since the ball is thrown horizontally, the time it takes to reach the ground is the same as the time it would take for an object to fall freely from the height of the building.

Using the formula for vertical displacement with the displacement as the height of the building, we can solve for time:

Height = 0 * Time + (1/2) * (-9.8) * Time^2

Simplifying the equation:

Height = -4.9 * Time^2

21.4 = -4.9 * Time^2

Dividing both sides by -4.9:

Time^2 = -21.4/-4.9

Time^2 = 4.3673469

Taking the square root of both sides:

Time = √4.3673469

Time ≈ 2.089 seconds

Now, we can substitute this time into the horizontal range formula to find the height of the building:

Range = Initial velocity * Time

21.4 = 13.0 * Time

21.4 = 13.0 * 2.089

21.4 = 27.157

Therefore, the height of the building is approximately 27.2 meters.

To determine the height of the building, we need to analyze the motion of the ball. Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. We can use the equation of motion to find the time taken for the ball to reach the ground:

d = v * t + (1/2) * a * t^2

In this case, the initial vertical velocity (v) is 0 m/s, the initial displacement (d) is the height of the building, the acceleration (a) is due to gravity (-9.8 m/s^2), and the time (t) is what we're trying to find.

Since the ball is thrown horizontally, the horizontal distance traveled by the ball is equal to the initial horizontal velocity (13.0 m/s) multiplied by the time (t):

distance = velocity * time
21.4 m = 13.0 m/s * t

Solving for time (t), we get:

t = 21.4 m / 13.0 m/s
t ≈ 1.646 s

Now that we have the time, we can use it to calculate the height of the building using the equation of motion:

d = v * t + (1/2) * a * t^2

Substituting the values, we get:

h = 0 * 1.646 + (1/2) * (-9.8) * (1.646)^2

Simplifying the equation:

h ≈ -4.9 * 1.646^2

Calculating the height of the building:

h ≈ -4.9 * 2.7078

h ≈ -13.272

Since height cannot be negative, we ignore the negative sign. Therefore, the height of the building is approximately 13.272 meters.