Two stones are thrown simultaneously, one straight upward with an initial speed

Question

Two stones are thrown simultaneously, one straight upward with an initial speed of 8.00 m/s from the base of a cliff and the other straight downward with an initial speed of 7.00 m/s from the top of the cliff that is 15.0 m above the base. When the stones cross paths, find (a) the time of flight, (b) the location (above the base), and (c) speed of the stone going up.
(a) 1.00 s; I got this part correct
(b) 3.1 m; I got his part correct
(c)   -2.9 m/s; I got this incorrect and I'm not sure if my significant figures are off or if it is a positive answer.

Explanation / Answer

(a)

for upward motion

y1 = v1*t + (1/2)*a*t^2

y1 = 8*t - (1/2)*9.8*t^2

for downward motion

-y2 = -v2*t - (1/2)*g*t^2

-y2 = -7t - (1/2)*9.8*t^2

y2 = 7t + (1/2)*9.8*t^2

but

y1 + y2 = 15

8t + 7t = 15

t = 1.00 s

------------

part (b)

y1 = 8*1 - (1/2)*9.8*1^2

y1 = 3.1 m

==============

part (c)

v = v1*t - g*t

v = 8*1 - 9.8*1

v = -1.80 m/s

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