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?

Answer 1

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.