A ball is launched from horizontal ground from the origin with an initial speed v_o and angle 45 $\circ$ above the horizontal.What is the horizontal position of the projectile when its horizontal position is equal to twice its vertical position? A)v_o²/g B)v_o²/2g C)3v_o²/2g D)2v_o²/g

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the kinematic equations of projectile motion. When the projectile reaches the height equal to half of its total vertical displacement (i.e., when y = h/2), the time of flight is t = v0sinθ/g. At this time, the horizontal displacement is given by x = v0cosθ*t. Substituting the expressions for t and x, we get x = v0^2sinθcosθ/g = v0^2/2g. Therefore, the answer is x = v0^2/2g = v_o²/2g, which is option B.

A Ball Is Launched from Horizontal Ground from the Origin ∘\circ∘

A Ball Is Launched from Horizontal Ground from the Origin $\circ$