An athlete at a Track and Field competition "puts" a shot put at 14m/s at a 40 degrees angle to the horziontal. How far from the athlete will the shot put land?
To find the horizontal distance the shot put will travel, we first need to find the horizontal component of the initial velocity. This can be calculated using trigonometry:
Horizontal component of velocity = initial velocity * cos(angle)
Horizontal component of velocity = 14m/s * cos(40 degrees)
Horizontal component of velocity ≈ 10.7 m/s
Now, we can use the horizontal component of velocity and the time the shot put is in the air to find the distance it will travel:
Time in the air can be calculated using the vertical motion equation:
Vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
0 = initial vertical velocity * time - (1/2) * 9.8 m/s^2 * time^2
Solving for time:
time = (initial vertical velocity) / (4.9 m/s^2)
time ≈ 1.43 sec
Now, we can find the horizontal distance the shot put will travel:
Horizontal distance = horizontal component of velocity * time
Horizontal distance = 10.7 m/s * 1.43 sec
Horizontal distance ≈ 15.3 meters
Therefore, the shot put will land approximately 15.3 meters from the athlete.