Ball A is tossed vertically upward from the top of a 40.0 m tall building with a

Q: Ball A is tossed vertically upward from the top of a 40.0 m tall building with a

a) t_A = voA/g = 20/9.8 = 2.04 s tB = voB/g = 30/9.8 = 3.06 s Hmax_B = voB^2/(2*g) = 30^2/(2*9.8) = 45.92 m b) let at time t they both are same height, let h is the height above the ground where they meet for ball B, h = VoB*t - 0.5*g*t^2 --(1) for ball A, -(40-h) = voA*t - 0.5*g*t^2 h = 40 + voA*t - 0.5*g*t^2 --(2) from equations 1 and 2 VoB*t - 0.5*g*t^2 = 40 + voA*t - 0.5*g*t^2 voB*t = 40 + voA*t t*(VoB - VoA) = 40 t = 40/(30-20) = 4 s

ID: 1406751 • Letter: B

Question

Ball A is tossed vertically upward from the top of a 40.0 m tall building with a speed of 20.0 m/s as shown in the figure below. At the same instant Ball B is tossed vertically upward from ground level with a speed of 30.0 m/s. Use the reference frame provided.

a) How long does it take for ball A to reach its maximum height?
time? How long does it take for ball B to reach its maximum height? How far above the ground is ball B at this time? (3 pts.)

b) Is there ever a time when both balls, A and B will be at the same height? If your answer is yes, then find the position and velocity for each ball, A and B at this time. If your answer is no, then find the time that each ball strikes the ground and find the velocity of each ball at this time. (4 pts.)

c) Sketch the position vs. time graph for ball A and ball B on the same plot.
Be sure to label your graphs and the relevant times and positions calculated in the above steps. (3 pts.)

Additional Practice: Find the maximum speed with which ball B could be tossed upward so that balls A and B will never be at the same height before reaching the ground level?

Explanation / Answer

a) t_A = voA/g

= 20/9.8

= 2.04 s

tB = voB/g

= 30/9.8

= 3.06 s

Hmax_B = voB^2/(2*g)

= 30^2/(2*9.8)

= 45.92 m

b) let at time t they both are same height,

let h is the height above the ground where they meet

for ball B,

h = VoB*t - 0.5*g*t^2 --(1)

for ball A,

-(40-h) = voA*t - 0.5*g*t^2

h = 40 + voA*t - 0.5*g*t^2 --(2)

from equations 1 and 2

VoB*t - 0.5*g*t^2 = 40 + voA*t - 0.5*g*t^2

voB*t = 40 + voA*t

t*(VoB - VoA) = 40

t = 40/(30-20)

= 4 s <<<<<----Answer

so they meet after 4 seconds.

position of ball A = 40 + 20*4 - 0.5*9.8*4^2

= 41.6 m from the ground

vA = voA - g*t

= 20 - 9.8*4

= 19.2 m/s

position of B = 41.6 from the ground

VB = VoB - g*t

= 30 - 9.8*4

= -9.2 m/s

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