two tall buildings are 80 m apart. A ball is thrown at 100 m/sec from one building toward the other but at an angle of 37 degrees below the horizontal. Find the vertical distance through which the ball falls before it hits the other building.

Question

two tall buildings are 80 m apart. A ball is thrown at 100 m/sec from one building toward the other but at an angle of 37 degrees below the horizontal. Find the vertical distance through which the ball falls before it hits the other building.

Answer 1

To find the vertical distance through which the ball falls before it hits the other building, we need to break down the ball's motion into horizontal and vertical components.

First, let's find the time it takes for the ball to reach the other building. We can use the horizontal distance and the horizontal component of the ball's velocity for this.

Given:
Distance between the buildings (horizontal distance) = 80 m
Horizontal component of velocity = 100 m/s (since the angle is below the horizontal, the vertical component does not contribute to the horizontal distance)

Using the equation: distance = speed * time (d = vt), we can find the time it takes for the ball to travel the horizontal distance:
80 m = 100 m/s * t
t = 80 m / 100 m/s
t = 0.8 s

Now that we know the time taken for the horizontal distance, we can find the vertical distance the ball falls during that time.

Given:
Vertical component of velocity = 37° below the horizontal
Acceleration due to gravity (g) = 9.8 m/s² (assuming no air resistance)

Using the formula for vertical motion: d = v₀t + (1/2)gt², where:
v₀ = initial vertical velocity (in this case, the vertical component of the ball's velocity)
t = time taken for the horizontal distance
g = acceleration due to gravity

First, let's find the initial vertical velocity (v₀):
To find the initial vertical velocity, we need to find the vertical component of the ball's velocity. We can use the trigonometric functions to determine this.

Given:
Magnitude of the velocity = 100 m/s
Angle below the horizontal = 37°

Vertical component of velocity (v₀) = magnitude of velocity * sin(angle)
v₀ = 100 m/s * sin(37°)
v₀ ≈ 100 m/s * 0.6018
v₀ ≈ 60.18 m/s

Now, we can substitute the values into the formula for vertical motion:
d = v₀t + (1/2)gt²
d = (60.18 m/s)(0.8 s) + (1/2)(9.8 m/s²)(0.8 s)²
d ≈ 48.144 m + (1/2)(9.8 m/s²)(0.64 s²)
d ≈ 48.144 m + 3.136 m
d ≈ 51.28 m

Therefore, the vertical distance through which the ball falls before hitting the other building is approximately 51.28 meters.