If a ball is thrown into the air with a velocity of 42 ft/s, its height (in feet) after t seconds is given by y = 42t − 16t2. find the velocity when t = 1.

The velocity when t = 1 is 10ft/s.

Given that,

Based on the above information, the velocity should be

v(t)=dy(t) ÷ dt

= d(42t - 16t^2) ÷ dt

= 42 - 32t

Since t=1

So, the velocity should be  

v = 42 - 32 × 1

= 10ft/s

Therefore we can conclude that the velocity when t = 1 is 10ft/s.

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the answer is 8.64

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