I have figured out the first balls velocity, time and etc. It\'s the second ball
ID: 1735951 • Letter: I
Question
I have figured out the first balls velocity, time and etc. It's the second ball that I seem to have a problem with. Icalculated the time it takes to travel up and stop, then the timeit takes to fall back down. I still get it wrong...Two students are on a balcony 19.2 mabove the street. One student throws a ball (ball 1) verticallydownward at 10.2 m/s; at the sameinstant, the other student throws a ball (ball 2) vertically upwardat the same speed. The second ball just misses the balcony on theway down. (a) What is the difference in the two ball'stime in the air?
1 s
(b) What is the velocity of each ball as it strikes the ground? ball 1 2 m/s ball 2 3 m/s
(c) How far apart are the balls 0.700s after they are thrown?
4 m (a) What is the difference in the two ball'stime in the air?
1 s
(b) What is the velocity of each ball as it strikes the ground? ball 1 2 m/s ball 2 3 m/s
(c) How far apart are the balls 0.700s after they are thrown?
4 m ball 1 2 m/s ball 2 3 m/s
Explanation / Answer
g=accleration due to gravity=9.8 time taken by the ball 2 to reach max. height=t let max height reached by the ball 2 =h v=u-g*t 0=10.2-9.8*t t=1.04 s v2=u2-2*g*h 0=10.2*10.2-2*9.8*h h=5.31 m let the ball 2 took T sec to reach the ground from the max.height H=5.31+19.2=24.51 H=u*t+(1/2)*g*T2 24.51=0+(1/2)*9.8*T2 T=2.24 total time the ball 2 was in the air was=t+T=1.04+2.24=3.28 s let the velocity of the ball 1 when it hit the ground was v v2=u2+2*g*h v2=10.2*10.2+2*9.8*19.2 v=21.92 (ans 2) the amount of time the ball 1 in air was t1 v=u+g*t1 21.92=10.2+9.8*t1 t1=1.2 s the difference in time of the two balls =3.28-1.2=2.08 s (ans1) let the velocity of the ball 2 as it hit the ground=v2 v22=u2+2*g*H v22=0+2*9.8*24.51 v2=21.92 m/s (ans 3) 0.7 s after the ball was thrown the ball 2 was going upwaeds let in 0.7 s the ball 2 has travelled x h=u*t-(1/2)*g*t2 h=10.2*0.7-(1/2)*9.8*0.7*0.7 h=4.739 m in 0.7 s the ball 1 travelled y m in downward direction h=u*t+(1/2)*g*t2 y=10.2*0.7+(1/2)*9.8*0.7*0.7 y=9.541 m after 0.7 s the 2 balls are 9.541+4.739 =14.28 m apart (ans4)