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# An arrow is shot off a cliff at an angle, with crosswind

Q: A student standing on the top of a cliff shoots an arrow from a height of 30.0 m at 25.0 m/s and an initial angle of 32.0Â° above the horizontal. There are four parts to this question.

a) What will be the horizontal and vertical components of the arrowâ€™s initial speed?

Horizontal component of initial velocity:
$V_{ix}=25\cdot cos(32)$
$V_{ix}=21.20 m/s$

Vertical component of initial velocity:
$V_{iy}=sin32\cdot 25.0$
$V_{iy}=13.25 m/s$

b) How high above the landscape under the cliff will the arrow rise? Assume a level landscape. (Hint: Find out how high it rises above the student. Then add 30.0 m.)

At the highest point, vertical velocity is zero. Initially, the vertical velocity is 13.25 m/s. As usual, acceleration (in the vertical) is equal to $-9.81 m/s^2$, due to gravity.

$V_{fy} = 0$
$V_{iy}=13.25 m/s$
$a=-9.81 m/s^2$
d= ?

Use the proper kinematics equation:

$v_f^2 = v_i^2 + 2ad$
$v_f^2 - v_i^2 = 2ad$

$d = \frac{v_f^2 - v_i^2}{2a}$

$d = \frac{0 - 13.25^2}{2(-9.81)}$

$d = \frac{-175.5625}{-19.62)}$

$d=8.95$

So the ball will rise 8.95 meters above the cliff, or (30+8.95) 38.95 meters above the ground.

c) How far off course will the arrow be after 3.35 s if there is a 12.0 m/s crosswind? (A crosswind is at 90Â° to the original motion; assume the arrow goes to the side at the speed of the crosswind.)

The crosswind is independent of the horizontal (x-axis) and vertical (y-axis) directions because it’s perpendicular to them. Think of the crosswind as the z-axis, coming out of the page.

So, how far it’s off-course is: d=vt = (12)(3.35)=40.2 m.

d) What will be the speed and angle of flight of the arrow 3.35 s after release if there is no crosswind?

To find its velocity at t=3.35 s, you have to find its vertical and horizontal velocities at that time, and then use pythagorean theorem to find the resultant velocity. At t=3.35 s:

Vx = 25 cos(32) = 21.20 m/s (it’s the same everywhere because there is no horizontal acceleration)

For the vertical, you know Vi, a, and t, and are looking for Vf. So use this equation:

$V_{fy} = V_{iy} + at$
$V_{fy} = 25*sin(32) + (-9.81)(3.35)$
$V_{fy} = -19.62 m/s$

It’s negative, which means it’s going downwards at t=3.35 s. This is the vertical component of the velocity. Next let’s find the resultant, or total velocity (the combination of horizontal and vertical components).

Draw a triangle:

Use pythagorean theorem to find that Vr = 28.88 m/s. This velocity triangle you drew is true only at t=3.35 s. It is constantly changing as the object moves.

For the angle, use trigonometry: $tan(\theta) = 19.62/21.20$

$\theta = 42.78$ degrees