A student is standing on the top of a building, and he throws an object up into the air at a velocity of 16.0 m/s at an angle of 25.0o to the horizontal. If the building is 75 m tall, how far from the base will the object hit the ground?
To solve this problem, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity can be calculated as:
Vx = V * cos(θ)
Vx = 16.0 m/s * cos(25.0o)
Vx = 14.53 m/s
The vertical component of the initial velocity can be calculated as:
Vy = V * sin(θ)
Vy = 16.0 m/s * sin(25.0o)
Vy = 6.77 m/s
Now, we can calculate the time it takes for the object to hit the ground using the vertical component of the initial velocity and the height of the building. We will use the following kinematic equation:
y = Vy * t + (1/2) * a * t^2
Where:
y = -75 m (height of the building)
Vy = 6.77 m/s (vertical component of initial velocity)
a = -9.81 m/s^2 (acceleration due to gravity, negative because it acts downwards)
-75 = 6.77 * t + (1/2) * (-9.81) * t^2
Solving the above quadratic equation gives us:
t ≈ 3.44 seconds
Now, we can calculate the horizontal distance traveled by the object using the horizontal component of the initial velocity and the time taken to hit the ground:
dx = Vx * t
dx = 14.53 m/s * 3.44 s
dx ≈ 50 meters
Therefore, the object will hit the ground approximately 50 meters away from the base of the building.