Problem: A person in a hospital bed sneezes. A droplet of saliva leavesthe perso

Question

Problem: A person in a hospital bed sneezes. A droplet of saliva leavesthe person's mouth at a velocity of 4.5 m/s at an angle of 45degrees to the horizontal and at a height of 1.5 m above the floor.How far away will the droplet land? I know there are the altered kinematic equations when theangle is known. I also have broken down the components ofthe triangle formed. I just can't seem to figure out what mynext step is. Thanks. Problem: A person in a hospital bed sneezes. A droplet of saliva leavesthe person's mouth at a velocity of 4.5 m/s at an angle of 45degrees to the horizontal and at a height of 1.5 m above the floor.How far away will the droplet land? I know there are the altered kinematic equations when theangle is known. I also have broken down the components ofthe triangle formed. I just can't seem to figure out what mynext step is. Thanks.

Explanation / Answer

velocity v = 4.5 m / s angle = 45 degrees height h = 1.5 m In vertical direction : -------------------- Initial velocity u = v sin = 3.181 m / s displacement S = h = 1.5 m Accleration a = 9.8 m / s^ 2 from the relation S = ut + ( 1/ 2) at ^ 2                    1.5 = 3.181 t + 4.9 t ^ 2 4.9 t^ 2 + 3.181 t - 1.5 = 0 t = { -3.181 ± [ 3.181^ 2 - ( 4*4.9*-1.5) ] } / 2( 4.9)     = { -3.181 ±6.286} / 9.8     = 0.316 s     sincetime is not negative In horizontal direction : ----------------------- velocity V = v cos 45 = 3.181 m / s Therefore required distance = V t                                            = 1 m

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