A shotputter throws the shot with an initial speed of 15.5 ms at a 34.0 angle to the horizontal calculate the horizontal distance traveled by the shot if it leaves the athletes hand at a height of 2?

Answer 1

-- The shot leaves his hand at 15.5 m/s directed 34 degrees above the horizon.

The horizontal component of its velocity is 15.5 cos(34) = about 12.85 m/s .

The vertical component of its velocity is 15.5 sin(34) = about 8.667 m/s .

-- The shot will continue to rise until its vertical velocity is zero. Since its vertical

acceleration is -9.807 m/s2 (g), it rises for 8.667/9.807 = about 0.884 second .

-- Its average speed during that time is (0.5 x 8.667) = about 4.333 m/s , and it

rises (0.884 x 4.333) = about 3.983 metersabove its release height, before it

starts falling. It then has 5.983 meters to fall before it hits the ground.

-- The time needed to fall 5.983 meters from the peak of its arc is

T = sqrt( 2h/g) = about 1.105 second. Added to the 0.884 second that it spent on the way up, that's a total of

(0.884 + 1.105) = about 1.989 seconds that it spends in the air after being

released.

-- Its horizontal progress in that time is (12.85) x (1.989) = about 25.55 meters .