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