Question

A ball is dropped from a tower 350 meters above the ground with position function s(t) = 4.9t2+ 350.

What is the velocity of the ball after 2 seconds?

Include units in your answer.

Answer 1

@CausticSyndicalist @kohai @ShadowLegendX @TheSmartOne @SolomonZelman

Answer 2

the equation is again

s(t)=4.9t^2+350

Answer 3

\(\large\color{slate}{ s(t) = 4.9t^2+ 350 }\)

velocity is given by \(\large\color{slate}{ s'(t) }\)

Answer 4

do we first find the derivative, so the velocity?

Answer 5

so you need \(\large\color{slate}{ s'(2) }\)

Answer 6

so the derivative of s(t)=4.9t^2+350 is s'(t)=9.8t

Answer 7

yes s'(t)=9.8t

Answer 8

so s'(2)=9.8(2)

Answer 9

which equals 19.6 ft/s

Answer 10

*m/s

Answer 11

meters not ft.

Answer 12

yes, it is distance/time, dimension is good:)
But it is meters.

Answer 13

yah i accidentally put feet by mistake

Answer 14

sure, but the main thing is that you understand:)
I am sure you do... yw

Answer 15

Thank you so much!

Answer 16

yw, 1s again!
Love working with knowledgeable questioners.

Answer 17

I will try to re-deserve my medals.
Remember that in in a position function \(\large\color{slate}{ s(t) }\)
(where s(t) is the position at time t)

the velocity function is always \(\large\color{slate}{ s'(t) }\).
(saying \(\large\color{slate}{ s'(t) }\) is \(\large\color{slate}{ v(t) }\) )

the acceleration function is \(\large\color{slate}{ s''(t) }\)
(saying \(\large\color{slate}{ s''(t) }\) is \(\large\color{slate}{ a(t) }\) )

Answer 18

(I can explain why if you would like to)

Answer 19

find the derivative of\[s(t)\] and then plug in 2 to find the velocity after 2 seconds\[s'(2)\]

Answer 20

Should the function be negative so the ball is falling rather than rising?