# don't understand?

A baseball player throws a baseball with the same initial velocity 5 different times. Each time the ball is thrown at a different angle. The angles are: 45 degrees, 15 degrees, 60 degrees, 25 degrees, and 80 degrees. (a) List the angles based on the amount of time in the air, from the shortest time to the longest time. (b) List the angles based on the horizontal range the baseball travels from the shortest distance to the longest distance.

### 4 Answers

- az_lenderLv 71 month agoFavorite Answer
(a) Steepest throws will spend the longest time in the air. 80, 60, 45, 25, 15.

(b) Range is (v^2/2g)*sin(2*theta), so just find out which angles lead to the largest values of "sin(2*theta)"

45, 60, 25, 15, 80.

- oubaasLv 71 month ago
flying time t = 2tup

tup = Vo*sin Θ /g =

the higher the sin , the higher the time t , the higher the angle (15, 25, 45, 60, 80)

range R ≡ sin 2Θ

sin 2*80° = sin 160° = sin 20° = 0.342

sin 2*15° = sin 30° = 0.5

sin 2*25° = sin 50° = 0.766

sin 2*60° = sin 120° = sin 60° = 0.866

sin 2*45° = sin 90° = 1.00

- AmyLv 71 month ago
Correct "velocity" to "speed." Assume no air resistance. Assume flat ground with the ball starting at ground level (I guess the pitcher is in a hole).

Time until the ball hits the ground depends only on the vertical motion. Rank the five throws by which has the largest vertical component.

Horizontal range depends on the horizontal component of velocity and on time in the air, and therefore on the vertical component. You can easily look up the formula or calculate the range individually for each throw.

- billrussell42Lv 71 month ago
calculate range and time for each angle, then sort them as requested. What don't you understand?

range of object projected at an angle θR = (V²/g)sin(2θ) V is initial velocity in m/s g is the acceleration of gravity 9.8 m/s²Time of flight = R/(Vcosθ)