A stone is thrown at 25 m / s at 53 o above the horizontal. At what time its velocity is at 45 o below the horizontal?

Step-by-step solution

Horizontal component of velocity, throughout the motion remain constant using u c o s θ = v c o s α 25 cos ⁡ 53 ° = v cos ⁡ 45 ° ⟹ v = 25 × 3 5 . 2 = 15 2 ⟹ v y = v sin ⁡ 45 = 15 Now using V y = u y + a y t i n y - direction, - 15 = 25 sin ⁡ 53° - g t - 15 =25 × 4 5 - 10 t ⟹ t = 3.5 sec