Example 2 A student throws a book horizontally out a dorm window with a speed of 12.5 m/s. The book lands on the ground 31.8 m from the base of the building. How high is the window above the ground?
We can solve this problem using the kinematic equations of motion. Since the book is thrown horizontally, its initial vertical velocity is zero, and the only force acting on it is gravity. We can use the following kinematic equation to find the height of the window above the ground:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the book to reach the ground.
To find the time, we can use the horizontal distance traveled by the book:
d = v_x * t
where d is the horizontal distance, v_x is the horizontal velocity (12.5 m/s), and t is the time.
Rearranging this equation gives us:
t = d / v_x
Substituting the given values:
t = 31.8 m / 12.5 m/s = 2.544 s
Now we can use the time to find the height:
h = (1/2) * g * t^2
h = (1/2) * 9.8 m/s^2 * (2.544 s)^2
h = 31.4 m
Therefore, the window is 31.4 meters above the ground
Answers & Comments
Answer:
31.4 meters
Explanation:
We can solve this problem using the kinematic equations of motion. Since the book is thrown horizontally, its initial vertical velocity is zero, and the only force acting on it is gravity. We can use the following kinematic equation to find the height of the window above the ground:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes for the book to reach the ground.
To find the time, we can use the horizontal distance traveled by the book:
d = v_x * t
where d is the horizontal distance, v_x is the horizontal velocity (12.5 m/s), and t is the time.
Rearranging this equation gives us:
t = d / v_x
Substituting the given values:
t = 31.8 m / 12.5 m/s = 2.544 s
Now we can use the time to find the height:
h = (1/2) * g * t^2
h = (1/2) * 9.8 m/s^2 * (2.544 s)^2
h = 31.4 m
Therefore, the window is 31.4 meters above the ground