If a ball is thown upwards with a velocity of 75 feet per second, what is its velocity at the end of 1.6 seconds??

In my calculator sheet) here's the data:

(g=32.2 ft/sec2)?
Velocity V = Vo - gl
= 75 - (32.2 x 16)
= 23.48 ft/sec.
Height S = Vol - 1/2 gl2 = (75x1.6)-(1/2 x 32.2 x (1.6)squared2)
= ?
thanks
redlight000

 

The velocity should be just 75 - (32.2 * 1.6) = 23.48

The distance travelled would be the mean of that velocity * 1.6 so
Mean of 75 & 23.48 = 49.24, * 1.6 = 78.784

That's how I would work it out, I hope it's correct..

 

This is a constant acceleration problem, so the usual kinematic equations apply. The key is to recognize that you know: the initial velocity, the acceleration (note it's negative), and the time. Distance as a function of time is a quadratic in time, so it's just plug and chug.

 

Hi Rjenkins,

well done Spot on.. congrats