Help me and i will rate u a lifesaver! Please give answer and explain using kine

Question

Help me and i will rate u a lifesaver! Please give answer and explain using kinematic equations!


Two students are on a balcony 19.5 m above the street. One student throws a ball vertically downward at 16.2 m/s; at the same instant the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in their time in air?
s
(b) What is the velocity of each ball as it strikes the ground?
m/s
m/s
(c) How far apart are the balls 0.700 s after they are thrown?
m

Explanation / Answer

the height of balcony s = 19.5m the velocity of ball moving down vi = 16.2m/s acceleration due to gravity g =9.8m/s^2 the final velocity when it reaches the ground vf =? from vf2=vi2+2gs vf2 =(16.2*16.2)+2*9.8*19.5 vf2=262.44+382.2 vf2=644.64 vf = 25.38m/s the time taken to reach the ground t1 =? from vf=vi+gt1 t1=(vf-vi)/g t1=(25.38-16.2)/9.8 t1=0.94sec. the time taken to the second ball to the same balcony t2= 2vi/g t2=2*16.2/9.8 t2=3.3sec the time taken for second ball to reach ground t=t1+t2 t=0.94+3.3=4.24sec a)so the difference of time between the two balls = t-t1=4.24-0.94=3.3sec b)the second ball also reaches the ground with the same velocity = 25.38m/s c)after 0.7sec the ball travels a distance s1=? from s1=vi*t+(1/2)gt2 s1=16.2*0.7+(1/2)*9.8*0.7*0.7 s1=11.34+2.4 s1=13.74m first ball moves 13.74m down the balcony and second ball is 13.74m above the balcony s1=16.2*0.7+(1/2)*9.8*0.7*0.7 s1=11.34+2.4 s1=13.74m first ball moves 13.74m down the balcony and second ball is 13.74m above the balcony

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