Kurt T. answered • 02/06/18

Math Tutoring and Test Prep

Assume the arrow is shot in a vacuum ... no wind, air resistance, etc.

We know the vertical component of the arrow's initial motion is 49 sin 30 = 24.5 m/sec.

We know the vertical component of the arrow's initial motion is 49 sin 30 = 24.5 m/sec.

We know the horizontal component of the arrow's initial motion is 49 cos 30 = 42.4 m/sec.

We know acceleration of gravity is -9.8 m/sec/sec.

The arrow will reach its maximum altitude in approximately 2.5 seconds. (24.5 / 9.8 = 2.5)

Assuming the arrow spends as much time falling as it did rising, it should be in the air for approximately 5 seconds.

Maximum height - attained 2.5 seconds after release - is 30.6 m, as you stated above.

Distance traveled is horizontal component times total flight time, or 5 * 49 cos 30 = 212.2 m.

Maximum height - attained 2.5 seconds after release - is 30.6 m, as you stated above.

Distance traveled is horizontal component times total flight time, or 5 * 49 cos 30 = 212.2 m.