Asked by tj
i posted this question before however i did not understand drwls explanation. anybody else want to try ?
Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?
Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?
Answers
Answered by
bobpursley
Well, it is easy if one considers energy.
if velocity doubles, then velocity quadruples
Vf^2=4Vi^2
but also we know from energy,
Vf^2=Vi^2+2*g*h where h is the distance fell.
so
4Vi^2=Vi^2+2gh
h= 3Vi^2/2g
Now how much time passes for it to go h?
average velocity= (Vf+Vi)/2= 3Vi/2
time= h/avg velocity= you do it. You know Vi
if velocity doubles, then velocity quadruples
Vf^2=4Vi^2
but also we know from energy,
Vf^2=Vi^2+2*g*h where h is the distance fell.
so
4Vi^2=Vi^2+2gh
h= 3Vi^2/2g
Now how much time passes for it to go h?
average velocity= (Vf+Vi)/2= 3Vi/2
time= h/avg velocity= you do it. You know Vi
Answered by
tj
does stand for gravity?
Answered by
bobpursley
g is the acceleration due to gravity, 9.8m/s^2
Answered by
tj
hmm i somehow got .3161 but that's wrong according to my online instructor.hmm
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