The same rock is thrown upwards at an angle of 45 degrees. It travels 10 meters up to its maximum height. How far will it travel before it lands?
To determine how far the rock will travel before it lands, we need to use the concept of projectile motion and consider the horizontal and vertical components of the rock's motion separately.
Given that the rock is thrown upwards at an angle of 45 degrees, we can split the initial velocity (V₀) of the rock into its horizontal (V₀x) and vertical (V₀y) components using trigonometric functions.
Since the angle is 45 degrees, both the horizontal and vertical components will have the same magnitude. Therefore, V₀x = V₀y = V₀/√2.
First, we need to find the time it takes for the rock to reach its maximum height. We can use the kinematic equation for vertical motion: Δy = V₀yt - (1/2)gt².
Since the rock travels 10 meters up to its maximum height and the vertical displacement (Δy) is zero at the highest point, we have:
0 = (V₀/√2)t - (1/2)gt².
Simplifying this equation, we get: t = (V₀/√2)/g.
To find the total time of flight, we can multiply the time it takes to reach the maximum height by 2, since the time taken to reach the maximum height is the same as the time taken to fall back down.
Therefore, the total time of flight is: T = 2 * [(V₀/√2)/g] = (2V₀/√2)/g = √2V₀/g.
Now, we can find the horizontal distance traveled by the rock before it lands using the equation: distance = horizontal velocity * time of flight.
The horizontal velocity (Vx) remains constant throughout the motion since there is no horizontal acceleration. So, Vx = V₀x = V₀/√2.
Therefore, the distance traveled before the rock lands is: distance = (V₀/√2) * √2V₀/g = (V₀²/g).
Given that the rock travels 10 meters up to its maximum height, we can deduce that (V₀/√2) * (V₀/√2)/g = 10.
Simplifying this equation, we get: (V₀²/g) = 20.
Thus, the rock will travel 20 meters before it lands.