**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:

Vertical component of initial velocity:

**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 , due to gravity.

d= ?

Use the proper kinematics equation:

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:

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).

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:

degrees

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