Saiquone+Selby

1) i.) 305 m/min

305/60= 5.08 m/s

(5.08 m/s)^2= 0 + 2(1.22 m/s^2)(x-xo)

25.8= 2.44(x-xo)

distance= 10.6 m

ii.) 190 m= (5.08 m/s)(t)- 1/2(-1.22 m/s^2)(t^2)

t= 13.9 s

5.08 m/s= 0+ (1.22 m/s^s)(t)

t=4.2 s

13.9 s+4.2 s= 18.1 s

2) i.) 50 m= Vo*t+ 1/2(9.8 m/s^2)(t^2)

50 m= 0+4.9(t^2)

t= 3.2 s ii.) 3.2-1= 2.2 s 2.2 s is the time of the second stone

50 m= Vo (2.2)+ 1/2(9.8 m/s^2)(2.2 s)^2

50 m= Vo (2.2) + 23.7

26.3 = 2.2 (Vo)

Vo = 11.9 m/s

Professor's Note : very Good effort - Give more explanation !! and try to avoid numerical values in the final answer -- check my page -

Nothing posted on for this week !!