You are standing on top of a building 135 meters tall. You throwa ball A upward

Question

You are standing on top of a building 135 meters tall. You throwa ball A upward with a velocity of 22.0 m/s. At the exact samemoment a friend throws ball B upward from the ground with velocityof 46.0 m/s. These two balls then collide at some later time.

               a) How long after these two balls are released will they collide?(answer: 5.63 s)

               b) Where will these two balls be when they collide? (answer: fromground=104 m)

               c) what will be the velocity of each ball just as theycollide?(answer: Ball a: -33.1 m/s, Ball b: -9.1m/s)

               d) What will be the relative velocity between these balls at themoment they collide? (answer: 24m/s)

WHAT ARE THE STEPS TO DOING THESE PROBLEMS? you shouldget the answers stated after each question.

Explanation / Answer

height of building H = 135 m After time t both balls are hit And height of the balls whenthey hitted be h Ball A : ------- Initial velocity u = 22 m / s distance S = H - h = 135 - h from the relation S = ut - ( 1/ 2) g t ^ 2                  -135+h = 22 t - (1/ 2) g t ^ 2                  -135+h= 22t - 4.9 t^ 2    -----( 1) ball B : ------- Initial velocity u ' = 46 m / s distance S ' = h fom the relation S ' = u't -( 1/ 2) gt ^ 2                       h= 46 t - 4.9 t^ 2                         h=46t -22t -135+h      since from eq (1)                         = 24 t - 135 + h                       0= 24t - 135                        t = 5.625 s plug this in eq ( 1 ) we get -135+h = 22(5.625)-4.9(5.625)^2                                                   h= 135 +(22*5.625) -4.9(5.625)^2                                                     = 103.71 m

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.