A stone is thrown from a rooftop at time t = 0 seconds. Its position at time t i
ID: 2857779 • Letter: A
Question
A stone is thrown from a rooftop at time t = 0 seconds. Its position at time t is given by r(t) = 3ti - 2tj + (9.8 - 4.9t^2) k. The origin is at the base of the building, which is standing on flat ground. Distance is measured in meters. The vector i points east, j points north, and k points up. (a) How high is the rooftop? meters. (b) When does the stone hit the ground? seconds. (c) Where does the stone hit the ground? (in meters) (d) How fast is the stone moving when it hits the ground? (in meters per second)Explanation / Answer
(a) height of rooftop
= k-component of r(0)
= 9.84.9(0)^2
= 9.8 meters
(b) Since the height is the k-component of r(t), the stone hits the ground when the k-component of r(t) equals 0.
9.8 - 4.9t^2 = 0
4.9t^2 = 9.8
t^2 = 2
t = + sqrt2
Use t = sqrt2 seconds since we don't want to use negative time in this context. The stone hits the ground in 2 seconds.
(c) The stone hits the ground at
r(2) = 3(sqrt2)i2(sqrt2)j+(9.84.9(sqrt2)^2)k = 3sqrt2 i - 2sqrt2 j.
So the stone hits the ground 3sqrt2 meters east and 2sqrt2 meters south of the base of the building.
(d) The stone's velocity vector at any time t is
v(t) = dr(t)/dt = 3i-2j-9.8tk.
When the stone hits the ground, its velocity vector is
v(2) = 8i-3j-9.8(sqrt2)k = 3i-2j-13.859k.
The speed is the magnitude (length) of the velocity vector, so when the stone hits the ground, the stone travels at a speed of
sqrt(3^2+(-2)^2+(-13.859)^2) = sqrt(205.072), or about 14.320, meters per second.
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