# Answer to Question #71318 in Mechanics | Relativity for Kashish

Question #71318

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? .

Expert's answer

Initial vector of velocity is

(v_0 cosθ,v_0 sinθ ).

Initial vector of velocity is

(v_0 cosθ,v_0 sinθ-gt).

It will be moving perpendicular to its initial direction of motion when scalar product of these vectors is zero:

v_0 cosθ (v_0 cosθ )+v_0 sinθ (v_0 sinθ-gt)=0

(v_0 cosθ )^2+(v_0 sinθ )^2=gtv_0 sinθ

gtv_0 sinθ=v_0^2

gt sinθ=v_0

t=v_0/(g sinθ ).

Answer:t=v_0/(g sinθ ).

(v_0 cosθ,v_0 sinθ ).

Initial vector of velocity is

(v_0 cosθ,v_0 sinθ-gt).

It will be moving perpendicular to its initial direction of motion when scalar product of these vectors is zero:

v_0 cosθ (v_0 cosθ )+v_0 sinθ (v_0 sinθ-gt)=0

(v_0 cosθ )^2+(v_0 sinθ )^2=gtv_0 sinθ

gtv_0 sinθ=v_0^2

gt sinθ=v_0

t=v_0/(g sinθ ).

Answer:t=v_0/(g sinθ ).

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