A confusing question for me. Help please

Question

I have answered the first question, but the second one is little confusing for me. 

 

Archery An arrow is shot at 30.0^o above the horizontal. Its velocity is 49m/s, and it hits the target.

 

a. What is the maximum height the arrow will attain?

Answer: (49sin30)²/2*9.8 = 30.625m

 

b. The target is at the height from which the arrow was shot. How far was it?

 

 

 

Answer 1

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 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.