A shot-putter throws the shot ( mass = 7.3 kg) with an initial speed of 15.0 m/s at a 33.0 degreCalculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 1.90 m above the ground.e angle to the horizontal.
To calculate the horizontal distance traveled by the shot put, you can use the equations of projectile motion. The horizontal distance is given by the formula:
d = v₀ * t * cosθ
Where:
- d is the horizontal distance traveled by the shot put.
- v₀ is the initial speed of the shot put.
- t is the time of flight.
- θ is the angle between the initial velocity vector and the horizontal direction.
To find the time of flight, you can use the equation:
t = 2 * (v₀ * sinθ) / g
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s²).
First, let's calculate the time of flight:
t = 2 * (15.0 m/s * sin(33.0°)) / 9.8 m/s²
t ≈ 1.82 seconds
Now, substitute the values into the first equation to find the horizontal distance traveled:
d = (15.0 m/s) * (1.82 s) * cos(33.0°)
d ≈ 23.85 meters
Therefore, the horizontal distance traveled by the shot put is approximately 23.85 meters.