- #1

Mesmer17

- 5

- 0

1.)A wild horse starts from rest and runs in a straight line 29 degrees north of west. After 34s of running in this direction, the horse has a speecd of 17m=m/s..

a.) north and east are positive lines, fine the component of the horses acceleration pointing along the north south line.

Here is what I did.. I found the average acceleration to be .5m/s^2.. then using the value .5 and the time 34s.. I plugged it into the equation Vxf=Vxo+AxTf.. so it would be Vxf=0+.5(34) which is 17..

b) find the component of the horses acceleration that points along the east west line..

I used the same equation but For Vo I used 17.. and used the rest of the numbers at before.. and I got..34.

Q2) When chasing a hare along flat ground,a grayhound leaps into the air at a speed of 12.3m/s at an angle 27degrees above the horizontal.

A) what is the range of his leap?

To do this I found, T setting 5.6(found from 12.3(sin27)) added to 1/2(-9.80)T equal to zero. The time comes out to 1.1s, then I used this multipiled by 12.3cos27.. to find the range... I just dont know if I did this right...

Q3) Car drives straight off a cliff 55m High, the impact of the car is 140m from the base of the cliff... how fast was the car traveling when it went over the cliff..

To do this one,I found the time by Using the Equation T=the sq.root of 2H/g.. so The sq.root of 2(55)/9.80.. to equal 3.4. then using the time, I used the equation Vyf=-g(T) so -9.80(3.4) which equals -33.3

PLEASE HELP ASAP.. LET ME KNOW WHAT I AM DOING WRONG..

THANKS