A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the horizontal. The stone hits tip of a structure which is 5 m tall, 6.1 s after being thrown. Ignore air resistance. (a) What is the height of the building?
(b) How far from the foot of the building does the stone land (only horizontal distance until the foot of the structure)?
(c) How fast is the stone moving just before it hits the ground?
(d) Please write down the acceleration vector in the unit vector notation for the stone at the instant it hits the ground.

The stone's acceleration, velocity, and position vectors at time are

where

and is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component of the position vector to 5 m and solve for :

(b) Evaluate the horizontal component of the position vector when :

(c) The rock's velocity vector has a constant horizontal component, so that

where

For the vertical component, recall the formula,

where and are the initial and final velocities, is the acceleration, and is the change in height.

When the rock hits the ground, it will have height . It's thrown from a height of , so . The rock is effectively in freefall, so . Solve for :

(where we took the negative square root because we know that points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

which has a magnitude of

(d) The acceleration vector stays constant throughout, so

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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