Chapter 2 Problem 66

Problem

A rock is thrown vertically upward with a speed of 12.0 m/s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 18.0 m/s.

(a) At what time will they strike each other?

(b) At what height will the collision occur?

(c) Answer (a) and (b) assuming that the order is reversed: the ball is thrown 1.00 s before the rock.

Solution

(a)

Choose up to be the positive direction. Let the throwing height of both objects be the $y$ = 0 location, and so $y_{0}$ = 0 for both objects. The acceleration of both objects is $a=g$.

The equation of motion for the rock:

$y_{\textrm {rock}}=y_{0}+v_{0{\textrm {rock}}}t+{\frac {1}{2}}at^{2}=v_{0{\textrm {rock}}}t+{\frac {1}{2}}gt^{2}$,

where $t$ is the time elapsed from the throwing of the rock.

The equation of motion for the ball, being thrown 1.00 s later, is

$y_{\textrm {ball}}=y_{0}+v_{0{\textrm {ball}}}(t-1.00{\textrm {s}})+{\frac {1}{2}}a(t-1.00{\textrm {s}})^{2}=v_{0{\textrm {ball}}}(t-1.00{\textrm {s}})+{\frac {1}{2}}g(t-1.00{\textrm {s}})^{2}$

Set the two equations equal(meaning the two objects are at the same place) and solve for the time of the collision.

$y_{\textrm {rock}}=y_{\textrm {ball}}\rightarrow v_{0{\textrm {rock}}}t+{\frac {1}{2}}gt^{2}=v_{0{\textrm {ball}}}(t-1.00{\textrm {s}})+{\frac {1}{2}}g(t-1.00{\textrm {s}})^{2}\rightarrow$

$\left(12.0{\textrm {m/s}}\right)t-{\frac {1}{2}}\left(9.80{\textrm {m/s}}^{2}\right)t^{2}=\left(18.0{\textrm {m/s}}\right)(t-1.00{\textrm {s}})-{\frac {1}{2}}\left(9.80{\textrm {m/s}}^{2}\right)(t-1.00{\textrm {s}})^{2}\rightarrow$

$\left(15.8{\textrm {m/s}}\right)t=\left(22.9{\textrm {m}}\right)\rightarrow t=1.45{\textrm {s}}$

(b)

Use the time of the collision in any trajectory $y_{\textrm {rock}}=v_{0{\textrm {rock}}}t+{\frac {1}{2}}gt^{2}=\left(12.0{\textrm {m/s}}\right)\left(1.45{\textrm {s}}\right)+{\frac {1}{2}}\left(9.80{\textrm {m/s}}^{2}\right)\left(1.45{\textrm {s}}\right)^{2}=7.10{\textrm {m}}$

(c)

Now the ball is thrown first so:

$y_{\textrm {ball}}=v_{0{\textrm {ball}}}t+{\frac {1}{2}}gt^{2}$

and

$y_{\textrm {rock}}=v_{0{\textrm {rock}}}(t-1.00{\textrm {s}})+{\frac {1}{2}}g(t-1.00{\textrm {s}})^{2}$

$y_{\textrm {rock}}=y_{\textrm {ball}}\rightarrow v_{0{\textrm {rock}}}(t-1.00{\textrm {s}})+{\frac {1}{2}}g(t-1.00{\textrm {s}})^{2}=v_{0{\textrm {ball}}}t+{\frac {1}{2}}gt^{2}\rightarrow$

$\left(3.8{\textrm {m/s}}\right)t=\left(16.9{\textrm {m}}\right)\rightarrow t=4.45{\textrm {s}}$

Does this answer make sense?

$y_{\textrm {ball}}=v_{0{\textrm {ball}}}t+{\frac {1}{2}}gt^{2}=\left(18.0{\textrm {m/s}}\right)\left(4.45{\textrm {s}}\right)+{\frac {1}{2}}\left(9.80{\textrm {m/s}}^{2}\right)\left(4.45{\textrm {s}}\right)^{2}=-16.9{\textrm {m}}$

The coordinate is below ground! So they never collide.