Problem 1A-6*: Vertically-Thrown Ball Without Air Resistance

A ball is thrown up vertically. The height of the ball, h(t), is given as a function of time t as

      h(t) = A + Bt + Ct2,

where the time t is taken to be zero at the moment that the ball is released.
The constant C is usually expressed as -g/2 with the meaning of the constant
g becoming clear later. Here we just use the name C.
A, B, and C are assigned the following values:

          A = 1.20 m,      B = 20.0 m/sec,       C = -4.90 m/sec2 .

(a) From which point is the ball released? How high will the ball rise?

(b) How long does it take for the ball to fall back to the ground?

(c) What is the speed when the ball strikes the ground?






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Solution:

(a) At t = 0, h(0) = A = 1.20 m since the ball is thrown from a point 1.20 m above the ground. The speed of the ball, v(t), is
      defined as

         

    When the ball reaches its maximum height at at time t, v(t) becomes zero.Therefore,
    by solving v(t) = 0, we will get the time t for the ball to reach its maximum height. Thus

         

    At t = 2.04 sec,

         h(2.04 sec) = (1.20 m)+(20.0 m/sec)×(2.04 sec)-(4.90 m/sec2)×(2.04 sec)2 = 21.6 m.

    Thus the ball reaches the maximum height of 21.6 m at t = 2.04 sec.

(b) Suppose at t=t1 the ball strikes the ground. To find t1 we need to solve h(t1)=0. Thus

         

   Since

         

   The negative solution t = -0.06 sec  is not acceptable; we must take t = 4.14 sec as
   the time when the ball strikes the ground. Thus it takes 4.14 sec for the ball to fall
   back to the ground.

(c) From (a) we know v(t) = B+2Ct. At t = 4.14 sec,


          v(4.14 sec) = (20.0 m/sec)+2×(-4.90 m/sec)×(4.14 sec) = -20.6 m/sec.

   Thus the ball strikes the ground with a speed of 20.6 m/sec. The minus sign in front
   of the answer means that the ball is moving downward when it hits the ground.

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