Answer to Question #71318 in Mechanics | Relativity for Kashish

Question

A projectile is thrown up with the initial speed u, making an angle theta (theta > 45 deg) with the horizontal. What is the time, just after which, it will be moving perpendicular to its initial direction of motion? .

Answer 1

Initial vector of velocity is
(v_0 cos⁡θ,v_0 sin⁡θ ).
Initial vector of velocity is
(v_0 cos⁡θ,v_0 sin⁡θ-gt).
It will be moving perpendicular to its initial direction of motion when scalar product of these vectors is zero:
v_0 cos⁡θ (v_0 cos⁡θ )+v_0 sin⁡θ (v_0 sin⁡θ-gt)=0
(v_0 cos⁡θ )^2+(v_0 sin⁡θ )^2=gtv_0 sin⁡θ
gtv_0 sin⁡θ=v_0^2
gt sin⁡θ=v_0
t=v_0/(g sin⁡θ ).
Answer:t=v_0/(g sin⁡θ ).