1010 Time to wait in dropping a stone | Rectilinear Translation

Problem 1010
A stone is thrown vertically up from the ground with a velocity of 300 ft per sec (91.44 m/s). How long must one wait before dropping a second stone from the top of a 600-ft (182.88-m) tower if the two stones are to pass each other 200 ft (60.96 m) from the top of the tower?

English System Solution

Click here to show or hide the solution

Stone from the ground:

$s = v_it + \frac{1}{2}at^2$

$h_1 = v_{i1}t - \frac{1}{2}gt^2$

$600 - 200 = 300t - \frac{1}{2}32.2t^2$

$16.1t^2 - 300t + 400 = 0$

$t = 17.19 \, \text{ sec and } \, 1.44 \, \text{ sec}$

Stone from the top of the tower:

Let t₂ = time to wait before dropping the second stone
$h = \frac{1}{2}g(t - t_2)^2$

With t = 17.19 sec
$200 = \frac{1}{2}(32.2)(17.19 - t_2)^2$

$t_2 = 13.67 \, \text{ sec}$

With t = 1.44 sec
$200 = \frac{1}{2}(32.2)(1.44 - t_2)^2$

$t_2 = -2.08 \, \text{ sec}$ (meaningless)

Use $t_2 = 13.67 \, \text{ sec}$ answer

System International Solution

Click here to show or hide the solution

Stone from the ground:

$s = v_it + \frac{1}{2}at^2$

$h_1 = v_{i1}t - \frac{1}{2}gt^2$

$182.88 - 60.96 = 91.44t - \frac{1}{2}9.81t^2$

$4.905t^2 - 91.44t + 121.92 = 0$

$t = 17.19 \, \text{ sec and } \, 1.44 \, \text{ sec}$

Stone from the top of the tower:

Let t₂ = time to wait before dropping the second stone
$h = \frac{1}{2}g(t - t_2)^2$

With t = 17.19 sec
$60.96 = \frac{1}{2}(9.81)(17.19 - t_2)^2$

$t_2 = 13.67 \, \text{ sec}$

With t = 1.44 sec
$60.96 = \frac{1}{2}(32.2)(1.44 - t_2)^2$

$t_2 = -2.08 \, \text{ sec}$ (meaningless)

Use $t_2 = 13.67 \, \text{ sec}$ answer

Tags:

translation

constant acceleration

free-falling body

stone thrown upward

stone drop from the tower

Rate:

No votes yet

Log in or register to post comments
26023 reads

1010 Time to wait in dropping a stone | Rectilinear Translation

Engineering Mechanics