Asked by Anonymous

a ball is thrown up vertically and caught at 9.3 sec. what is the intial velocity
Answered by bobpursley
hf=hi+vi*t- 4.9 t^2
0=0+Vi*9.3-4.9*9.3^2 solve for vi
Answered by Anonymous
is the answer 386.556??
Answered by bobpursley
No. You can do better than that.
Answered by Anonymous
whats the answer?
Answered by bobpursley
You have come to the wrong place to mooch answers. You learn nothing doing that.
Solve this: 0=Vi*9.3-4.9*9.3^2
Answered by Anonymous
423.801?
Answered by Steve
i got the same question, however how would you determine the maximum hieght it reaches?
Answered by bobpursley
The time to get to the top is 1/2 * 9.3sec

hf=ho+Vi*(1/2*9.3)-4.9(1/2*9.3)^2

ho=0, vi you solved for, find hf.

No, anonomus.

vi*9.3= 4.9*9.3^2
surely you can solve this equation.
It is less than 50m/s
Answered by Anonymous
45.57!
Answered by Steve
thankyou bob. i calculated the problem and got roughly 106 m.
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