1. An object is launched from a known height at a known speed and a known angle

Question

1. An object is launched from a known height at a known speed and a known angle above the horizontal. Which single equation allows you to find the vertical component of its velocity vector just before it hits the ground? ( i got y= yo+voyt+1/2ayt^2 and its wrong.)  choose from:

x= xo+voxt+1/2axt^2

vx= vox+axt

vx^2= vox^2+ 2ax (x-xo)

vy= voy+ayt

vy^2= voy^2+ 2ay(y-yo)

y= yo+voyt+1/2ayt^2

2 .A roach that runs for its life because you try to hit it with your physics book runs off the horizontal tabletop with a speed of 0.546 m/s. It strikes the floor a horizontal distance 0.393 m away from the edge of the table. Ignore air resistance.
(Show the individual steps of your work. Just the right answer will not give full credit.)
a) What will be the horizontal component of the roach's velocity just before it hits the floor? (I got -7 m/s and it was wrong)

b) what is the height of the table?

c) What will be the vertical component of the roach's velocity just before it hits the floor?

3. An object is launched at a known speed and a known angle with respect to the horizintal. Check all questions that simply state that the horizontal component of the velocity vector does not change. (Ignore air resistance.) (I got vx^2=vox^2+2ax(x-xo) and vy^2=voy^2+2ay(y-yo) and it was wrong) choose from options of first problem.

4. A baseball is thrown horizontally at a speed of 40.9 m/s. When it reaches home plate, what is the horizontal component of its velocity?

5. A baseball is thrown with 31.4 m/s at an angle of 22.1 degree above the horizontal. It is caught at the same height by another player a distance of 32 yards away. What is the speed of the ball, just before it is caught?

Explanation / Answer

1)vy^2= voy^2+ 2ay(y-yo)

2) horizontal component doesnt change so 0.546 m/s

b) first x direction
x = v0x t

t = 0.393/0.546=0.72 s

y = y0 + v0y t + 1/2 a t^2

0 = h - 0.5*9.81*0.72^2
h=2.54 m

c) vy = v0 + at = -9.81*0.72=-7.06 m/s

3)

vx= vox+axt

and
vx^2= vox^2+ 2ax (x-xo)

4) horizontal component doesnt change so 40.9 m/s

5) since same height speed doesnt change
31.4 m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.