Question

part 1 of 3
let ( h = s(t) = 50t - 16t^{2} ) give the height of the ball at time ( t ).
then the balls velocity at time ( t = a ) can be found by
( v(a)=lim_{t
ightarrow a}\frac{s(t)-s(a)}{t - a} ).
we are requested to find the velocity at ( t = 2 ); therefore we use ( a = 2 ) and have
( v(2)=lim_{t
ightarrow 2}\frac{s(t)-s(2)}{t - 2} )
( =lim_{t
ightarrow 2}\frac{(50t - 16t^{2})-(50(quad)-16(2)^{2})}{t - 2} )
( =lim_{t
ightarrow 2}\frac{(50t - 16t^{2})-36}{t - 2} ).

Explanation:

Step1: Substitute \(a = 2\) into \(s(a)\)

We know \(s(t)=50t - 16t^{2}\), so \(s(2)=50\times2-16\times2^{2}\).

Step2: Calculate \(s(2)\)

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Answer:

The value to fill in the blank is \(2\) (from \(s(a)\) with \(a = 2\) in the first - level blank).