Physics Question

Question

Two students are on a balcony 18.3 m above the street. One student throws a ball, b₁, vertically downward at 13.3 m/s. At the same instant, the other student throws a ball, b₂, vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a)  What is the difference in time the balls spend in the air?

 ________

(b) What is the velocity of each ball as it strikes the ground?

 

velocity for b₁   = ________ m/s

velocity for b3 = _________ m/s

(c) How far apart are the balls 0.490 s after they are thrown?

 ______ m

 

Answer 1

Part a: To figure out the difference in time, it's as simple as calculating how long the second ball (b2) will reach back to the balcony's height after being thrown upward. Treating the balcony as zero height, we can use the following kinematic equation:

y=y₀+v₀t-(1/2)gt²

where y=y₀=0 because the starting and final height of b₂ is at the balcony, g=9.8 m/s², and v₀=13.3 m/s. After plugging in the numbers, equation becomes

0=13.3t-(1/2)(9.8)t²

0=t(13.3-4.9t)

Solving will get us t=0 and t=2.71 s. From there, 't = 2.71 seconds' is your answer.

Part b: Both balls should have the same velocities when they strike the ground. Although b₂ started with an upward velocity of 13.3 m/s, it'll have a downward velocity of 13.3 m/s (same as b₁) when it reaches back at the balcony's height. Use the following to calculate the final velocities using b₁:

v²=v₀²-2g(y-y₀)

where v₀=-13.3 m/s, y=0, and y₀=18.3m. Here, I'm setting the ground to be zero height. Plug in the numbers

v²=(-13.3)²-2(9.8)(0-18.3)

v=23.09m/s

The answer should be v=-23.09 m/s where the negative accounts that the balls are traveling downward.

Part c: We can use the following equation for both balls

y=y₀+v₀t-(1/2)gt² where we set the ground as zero height. Here y₀=18.3m. Also t =0.49s

For b₁, we have

y₁=18.3+(-13.3)(0.49)-(1/2)(9.8)(0.49)²

y₁=10.61m

For b₂, we have

y₂=18.3+(13.3)(0.49)-(1/2)(9.8)(0.49)²

y₂=23.64 m

Take the difference between the two ball's height

y₂-y₁= 23.64 - 10.61 = 13.03 m

That's it, you're done!

Answer 2

h = 18.3 m but treat this as the 0 point

vb1 = -13.3 m/s; vb2 = 13.3 m/s

a) b1: -18.3 = 0 - 13.3t - (0.5)(9.8)t² ; use quadratic formula to find t = 1.0 s

b2: -18.3 = 0 + 13.3t - (0.5)(9.8)t² ; use quadratic formula to find t = 3.7 s; 3.7 - 1.0 = 2.7 s

b) The ball thrown upward will return to the same throw position with the same 13.3 m/s but downward. Because that is the same downward velocity as the ball that was thrown downward. So they will hit the ground at the same velocity.

c) At 0.49 s: hb1:= 0 - 13.3(0.49) - ((0.5)(9.8)(0.49)² = -7.69 m (below the balcony)

hb2:= 0 + 13.3(0.49) - ((0.5)(9.8)(0.49)² = 5.34 m (above the balcony)

Thus they were 13.0 m apart