A stone is thrown downward off cliff with initial speed of v i and angle of ? fr

Question

A stone is thrown downward off cliff with initial speed of v i and angle of ? from vertical. The rock strikes the ground t seconds later. 1) What is height of cliff 2) What is distance from base of cliff at which stone lands 3) What is vertical speed of stone at impact 4) What is net speed of stone at impact? A stone is thrown downward off cliff with initial speed of v i and angle of ? from vertical. The rock strikes the ground t seconds later. 1) What is height of cliff 2) What is distance from base of cliff at which stone lands 3) What is vertical speed of stone at impact 4) What is net speed of stone at impact?

Explanation / Answer

vertical initial velocity = Vi sin ? where ? = angle of projection time to reach the ground = t applying kinematics equation in vertical direction :----
v(y) final velocity in y direction
u(y) initial velocity in y- direction = Vi sin ? acceleration in y- direction = g H = height of cliff = u(y)*t + (1/2)*g*t^2 So; height of cliff = Vi (sin ?) t + (g/2)t^2 -------------------------------Ans applying kinematics equation in vertical direction :----
v(x) final velocity in x direction
u(x) initial velocity in x- direction = Vi cos ? acceleration in x- direction = a=0 R =Horizontal range =  u(x)*t + (1/2)*a*t^2 distance from base of cliff at which stone lands = Vi (cos ?)*t ---------------------------------Ans applying kinematics equation in vertical direction :---- V(y)= u(y) + a*t vertical velocity at the time of impact=V(y)= Vi sin ? +gt-----------------------------Ans applying kinematics equation in horizontal direction :---- V(x)= u(x) + a*t V(x) = u(x)+0*t horizontal velocity at the time of impact=V(x)= Vi cos ? so; net speed of stone at impact = [ (v(x))^2 + (v(y))^2 ]^(1/2)    = [ (v(i))^2 + (gt)^2 + 2Vi * g*t*sin ?]^(1/2) ---------------Ans
v(x) final velocity in x direction
u(x) initial velocity in x- direction = Vi cos ? acceleration in x- direction = a=0 R =Horizontal range =  u(x)*t + (1/2)*a*t^2 distance from base of cliff at which stone lands = Vi (cos ?)*t ---------------------------------Ans applying kinematics equation in vertical direction :---- V(y)= u(y) + a*t vertical velocity at the time of impact=V(y)= Vi sin ? +gt-----------------------------Ans applying kinematics equation in horizontal direction :---- V(x)= u(x) + a*t V(x) = u(x)+0*t horizontal velocity at the time of impact=V(x)= Vi cos ? so; net speed of stone at impact = [ (v(x))^2 + (v(y))^2 ]^(1/2)    = [ (v(i))^2 + (gt)^2 + 2Vi * g*t*sin ?]^(1/2) ---------------Ans

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