a ball is projected with a speed of 20 m/s from a building toward a second building as shown. ignore air resistance and take g = 10 m/s. where does the ball hit and how long is it in the air if the initial angle is 45 degrees with the horizontal?
You neglected to say how far away the second building is. That information is required.
To determine where the ball hits and how long it stays in the air, we can use the equations of motion under projectile motion.
First, we need to break down the initial velocity of the ball into its horizontal and vertical components. The initial angle of 45 degrees with the horizontal will give us equal magnitudes for these components.
The horizontal component (Vx) can be calculated using the equation Vx = V * cosθ, where V is the magnitude of the initial velocity (20 m/s) and θ is the angle (45 degrees).
Vx = 20 m/s * cos(45°)
Vx = 20 m/s * √2 / 2
Vx = 10√2 m/s
The vertical component (Vy) can be calculated using the equation Vy = V * sinθ.
Vy = 20 m/s * sin(45°)
Vy = 20 m/s * √2 / 2
Vy = 10√2 m/s
Next, we can use the vertical component to find the time of flight (t) by using the equation:
t = 2 * Vy / g
Substituting the values:
t = 2 * (10√2 m/s) / 10 m/s^2
t = 2√2 s
The time of flight is approximately 2.83 seconds.
Now, to determine where the ball hits, we can use the horizontal component:
x = Vx * t
Substituting the values:
x = (10√2 m/s) * (2√2 s)
x = 40 m
Therefore, the ball hits a distance of 40 meters from the building it was projected from.