Asked by Jin
If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, it height in feet after t second is given by y=95tâ16t2. Find the average velocity for the time period begining when t=1 and lasting
(i) 01 seconds:
(ii) 001 seconds:
(iii) 0001 seconds:
Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1.
(i) 01 seconds:
(ii) 001 seconds:
(iii) 0001 seconds:
Finally based on the above results, guess what the instantaneous velocity of the ball is when t=1.
Answers
Answered by
Reiny
i)
when t=1, y = 95-16 = 79
when t=1.01 , y = 95(1.01) - 16(1.01)^2 = 79.6284
avg vel = (79.6284-79)/(1.01-1) = 62.84
repeat for ii) and iii)
you should be able to draw your conclusion by looking at those results.
when t=1, y = 95-16 = 79
when t=1.01 , y = 95(1.01) - 16(1.01)^2 = 79.6284
avg vel = (79.6284-79)/(1.01-1) = 62.84
repeat for ii) and iii)
you should be able to draw your conclusion by looking at those results.
Answered by
bill
When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where âvâ is the init
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