Phy231

Seed 8.2

A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an

approximation within 2% of the 9.8 m/s^2 acceleration of gravity).

How high does it rise and how long does it take to get to its highest point?

answer/question/discussion: ->->->->->->->->->->->-> :

v0=15m/s a=-10m/s^2 vf=0m/s 'dt=(0m/s+15m/s)/a=+-1.5s 'ds=(vf+v0)/2*'dt 'ds=(0m/s+15m/s)/2*'1.5s 'ds=11.25m

#$&*

How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?

answer/question/discussion: ->->->->->->->->->->->-> :

v0=0m/s 'ds=11.25m+12m=22.25m a=-10m/s^2 vf^2=v0^2+2a'ds vf=+-'sqrt(v0^2+2a'ds)=+-'sqrt(0^2+2*-10*22.25)=+-'sqrt(445)=+-21.1m/s

'dv=0-(-21.1m/s)=21.1m/s 'dt=21.1m/s/a=21/1m/s/-10m/s^2=2.11s

#$&*

At what clock time(s) will the speed of the ball be 5 meters / second?

answer/question/discussion: ->->->->->->->->->->->-> :

vf=5m/s v0=15m/s a=-10m?s^2 'dt=(5m/s+15m/s)/a=+-2s

#$&*

At what clock time(s) will the ball be 20 meters above the ground?

How high will it be at the end of the sixth second?

answer/question/discussion: ->->->->->->->->->->->-> :

'ds=20m v0=15m/s a=-10m/s^2 vf^2=v0^2+2a'ds vf=+-'sqrt(v0^2+2a'ds)=+-'sqrt(15^2+2*-10*20)=+-13.23m/s

'ds=(15m/s+13.23m/s)/2*6s=84.69m If you do negative then it is 'ds=(15m/s+(-13.23m/s)/2*6s=5.31m

'dt=(15m/s+13.23m/s)/-10m/s^2=2.8s

25min

3/8/10 12pm