Asked by austin
Standing on a balcony, you throw your keys to a friend standing on the ground below. 1.1 seconds after you release the keys, they have an instantaneous velocity of 15.8 m/s, directed 60° below the horizontal. What initial velocity did you give them? magnitude:
direction: below the horizon
direction: below the horizon
Answers
Answered by
bobpursley
Well, the horizontal component is constant.
Horizontal V=15.8*cos60= you do it.
Now in the vertical, vf(1.1)=Viv+gt this assumes the keys were thrown downward, there is another solution..
15.8*sin60=Viv+9.8(1.1)
solve for Viv
then, intial velocity= horizonal in x + verticalinitial in y
magnitude= sqrt(hor^2+viv^2)
angledownward= arctan downward/horizontal
Horizontal V=15.8*cos60= you do it.
Now in the vertical, vf(1.1)=Viv+gt this assumes the keys were thrown downward, there is another solution..
15.8*sin60=Viv+9.8(1.1)
solve for Viv
then, intial velocity= horizonal in x + verticalinitial in y
magnitude= sqrt(hor^2+viv^2)
angledownward= arctan downward/horizontal
Answered by
austin
but what would the magnitude be and also the direction in degrees
Answered by
bobpursley
my last two lines show you how to compute. What do you not understand about those lines?
Answered by
austin
i don't understand what numbers yu plug into this:
magnitude= sqrt(hor^2+viv^2)
angledownward= arctan downward/horizontal
magnitude= sqrt(hor^2+viv^2)
angledownward= arctan downward/horizontal
Answered by
bobpursley
hor= horizontal component found first.
viv= initial vertical velocty, from 15.8*sin60=Viv+9.8(1.1)
arc tan means the angle whose tangent is (vertical velocity initial/horizontal initial)
viv= initial vertical velocty, from 15.8*sin60=Viv+9.8(1.1)
arc tan means the angle whose tangent is (vertical velocity initial/horizontal initial)
Answered by
austin
i don't understand how to get the degree part of this question