Question

Two physics students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s; at the same instant, the other student throws a ball vertically upward at the same speed. The second balljust misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b)What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls 0.8 s after they are thrown?

Answer 1

you need to understand position and velocity equation for this problem
\[x(t)=\frac{1}{2}at^2+v_ot+x_o\]
\[v(t)=at+v_o\]

now for the general constants
acceleration (a) is equal to acceleration of gravity(g) =-9.81 m/s^2
the initial position x_o =19.6m

Answer 2

now 2 cases
if you throw the ball upwards, then v_o= 14.7
if you throw the ball downwards, then v_o=-14.7

to determine a)
substitute the values into the position equation and determine t when x(t)=0 for both cases
x(t)=0 means when the balls reach the ground or hit the ground and the corresponding t value would be the time it reaches the ground

now the difference of the 2 times would be your answer for a

Answer 3

now for b)
once you have the 2 t values for once they reach the ground

substitute the corresponding values into the velocity equation given above to determine the velocity at which it hits the ground for each of the time values
basically solve v(t) for both t values

Answer 4

c)
for the 2 cases
substitute the respective values into the position equation, then determine x(.8)
x(.8) is the y position of the ball when the time is equal to .8 seconds
then find the difference between the 2 position values

Answer 5

@K0GA !! :D

Answer 6

yea :)

Answer 7

perfect completeidiot

Answer 8

i can't understand what you want from a)
tell me what he means please :)

Answer 9

answer for b) the two balls have the same velocity = 24.5 m\s

Answer 10

basically solving for x(t)=0=.5at^2+v_ot+x_o for t

Answer 11

for both instances and then finding the difference in time

Answer 12

time 1 is 4? sorry...i'm so slow...i can't find my scientific calculator...i'm such an idiot..:(

Answer 13

answer c) is 23.49 m

Answer 14

how to compute for time 2?
\[0=(-14.7)t _{2}-(.5)(9.8)^2+0\]
@completeidiot ?

Answer 15

quadratic equation
and x_o is not equal to 0 because you're throw it off a high ledge with certain height
also you're missing t_2 next to acceleration

Answer 16

\[-\frac{1}{2}9.8t_2^2-14.7t_2+19.6=0\]

Answer 17

oh..ok..

Answer 18

t_2 = 1?
t_1= 4?

Answer 19

@K0GA am i correct?

Answer 20

yea u r perfect

Answer 21

but there are another solution shorter

Answer 22

because There is no air resistance

Answer 23

so Difference in time will be Time required up and down to balcony

Answer 24

you got it ?!

Answer 25

ok?

Answer 26

you do you mean by ok?!

Answer 27

i got what you're saying...

Answer 28

if we neglect air resistance

Answer 29

when i throw the first ball up it will go up until rest then we go down right ?!

Answer 30

when it arrive the same place its velocity will be the same

Answer 31

right ?!

Answer 32

yeA

Answer 33

ok so when it get back to the same place with the same velocity down it will be like the second stone

Answer 34

so the difference in time will be the time required to go up and down to balcony

Answer 35

\[v=v0+at\]

Answer 36

\[-14.7=14.7-9.8t\]

Answer 37

\[9.8t=29.4\]

Answer 38

wat's the answer to that? sorry...just lost my calcu...

Answer 39

\[t=3 \sec\]

Answer 40

a) 3 sec

Answer 41

is that your answer ?!

Answer 42

wait...from idiot's solution...i got 4 :(

Answer 43

you got 4 sec for first stone ?!

Answer 44

yes..

Answer 45

and 1 sec for first stone . right !

Answer 46

so time difference is 3 sec

Answer 47

ahh...you're good! :)

Answer 48

you understand what i said ?

Answer 49

yes...but is that solution enough? for letter a?