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Chapter 3: Problem 64

Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed \(v_{0}\) from height \(h\)

Short Answer

Expert verified
The general formula for the horizontal distance covered by a projectile launched horizontally from height \(h\) at speed \(v_{0}\) is \(d = v_{0}\sqrt{2h / g}\).

Step by step solution

01

Derive the time of flight

Firstly, you need to figure out the time it takes for the projectile to hit the ground (this is often referred to as the time of flight). Since the projectile is launched horizontally, initial vertical velocity (\(v_{y_{0}}\)) is 0. We can use the equation of motion to find the time (\(t\)) it takes for a body to fall from a height \(h\) under gravity \(g\). The equation of motion is \(h = v_{y_{0}}t + 0.5gt^{2}\), by substituting \(v_{y_{0}}=0\) into the equation we get: \(h = 0.5gt^{2}\). Solving this for \(t\) gives us the formula for time of flight: \(t = \sqrt{2h / g}\).
02

Calculate the horizontal distance

Next, use the formula of horizontal distance (d) traveled by a projectile, which is the product of time of flight (\(t\)) and horizontal speed (\(v_{0}\)). Substitute the value of \(t\) from the previous step into the equation: \(d = v_{0}t\). Thus, substituting \(t = \sqrt{2h / g}\) we get the formula: \(d = v_{0}\sqrt{2h / g}\).
03

General formula

So, the general formula for the horizontal distance covered by a projectile launched horizontally at speed \(v_{0}\) from height \(h\) is \(d = v_{0}\sqrt{2h / g}\).

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