A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case, and air resistance is negligible. Case A: Thrown straight up. Case B: Thrown straight down. Case C: Thrown out at an angle of 45° above horizontal. Case D: Thrown straight out horizontally. In which case will the speed of the stone be greatest when it hits the water below?

Since Case D has the highest value for v₁ = √ (u² + 2gs). The stone has the greastest speed in this case.

Explanation:

Let u be the speed of the stone

Case A - If the stone is thrown straight up, using v² = u² - 2gs, where s is the height of the bridge above the ground. The speed as it hits the ground is v₁ = √2gs

Case B - If it is thrown straight down, using v² = u² - 2gs, where s is the height of the bridge above the ground. The speed as it hits the ground is v₁ = √u² - 2gs

Case C - If the stone is thrown at an angle of 45° to the horizontal, we have both horizontal component ucos45 and vertical component usin45. Using v² = u² - 2gs, where s is the height of the bridge above the ground. The vertical speed as it hits the ground is v = √{ (usin45) ² - 2gs]. Thus the speed as it hits the ground is thus v₁ = √[v² + (ucos45) ²] = √[ (usin45) ² - 2gs] + (ucos45) ²] = √[ (usin45) ² + (ucos45) ² - 2gs] = √ (1 - 2gs)

Case D - If the stone is thrown out horizontally, we have both horizontal component u and initial vertical component v = 0. Using v² = u² - 2gs, where s is the height of the bridge above the ground. The vertical speed as it hits the ground is v = √{0² - 2g * - s]. = √2gs Thus the speed as it hits the ground is thus v₁ = √[v² + u²] = √ (u² + 2gs)

Since Case D has the highest value for v₁ = √ (u² + 2gs). The stone has the greastest speed in this case.