A shotputter throws the shot with an initial speed of 16.0 m/s at a 43.0 angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 m above the ground.
To calculate the horizontal distance traveled by the shot, we can break down the initial velocity of the shot into horizontal and vertical components.
The initial vertical velocity (Vy) is calculated using the formula:
Vy = V * sinθ
where:
V = initial speed = 16.0 m/s
θ = angle of projection = 43.0 degrees
Vy = 16.0 m/s * sin(43.0°)
= 16.0 m/s * 0.68199836006
≈ 10.912 m/s
The time of flight (t) can be calculated using the equation:
t = (2 * Vy) / g
where:
g = acceleration due to gravity = 9.8 m/s²
t = (2 * 10.912 m/s) / 9.8 m/s²
≈ 2.222 seconds
Next, we can calculate the horizontal distance (dx) using the formula:
dx = Vx * t
where:
Vx = initial horizontal velocity
t = time of flight
To find Vx, we can use the formula:
Vx = V * cosθ
Vx = 16.0 m/s * cos(43.0°)
= 16.0 m/s * 0.75390225434
≈ 12.063 m/s
Now, we can calculate the horizontal distance:
dx = 12.063 m/s * 2.222 s
≈ 26.835 meters
So, the shot will travel approximately 26.835 meters horizontally.