Asked by dan
Sam tosses a ball horizontally off a footbridge at 3.2 m/s. How much time passes after he releases it until its speed doubles?
Answers
Answered by
drwls
The vertical velocity component Vy would have to increase to sqrt3 times the initial horizontal velocity, Vx, so that V^2 = Vx^2 + Vy^2 = 4 Vx^2, and
V = 2 Vx
To achive that velocity, it must fall a distance H such that
M g H = (1/2) M Vy^2 = 3/2 M Vx^2
H = [3/(2g)]*Vx^2
V = 2 Vx
To achive that velocity, it must fall a distance H such that
M g H = (1/2) M Vy^2 = 3/2 M Vx^2
H = [3/(2g)]*Vx^2
Answered by
marco
i don't understand your work. please try to explain it more thoroughly.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.