A shotputter throws the shot with an initial speed of 15.6 m/s at a 38.0 angle to the horizontal.
Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.15 m above the ground.
To calculate the horizontal distance traveled by the shot, we can use the principles of projectile motion.
First, let's break down the initial velocity of the shot into its horizontal and vertical components.
The horizontal component of the velocity (Vx) remains constant throughout the motion since there are no horizontal forces acting on the shot. So, Vx = 15.6 m/s.
The vertical component of the velocity (Vy) can be calculated using the initial speed and the angle of projection.
Vy = V₀ * sin(θ)
= 15.6 m/s * sin(38.0°)
≈ 9.6 m/s
Now, we can determine the time it takes for the shot to hit the ground. Since the vertical motion is affected by gravity, we can use the equation:
y = y₀ + V₀y * t + (1/2) * (-g) * t²
where y₀ is the initial vertical position (2.15 m above the ground), V₀y is the initial vertical velocity (-9.6 m/s), t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Setting y = 0 (since the shot will hit the ground), we can solve for t:
0 = 2.15 m + (-9.6 m/s) * t + (1/2) * (-9.8 m/s²) * t²
Simplifying the equation, we have:
4.9 m/s² * t² + 9.6 m/s * t - 2.15 m = 0
We can solve this quadratic equation to find the value of t.
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where a = 4.9 m/s², b = 9.6 m/s, and c = -2.15 m, we can calculate t.
Using the positive value for t (ignoring the negative value because time cannot be negative in this case), we find:
t ≈ 1.6 s
Now, we can calculate the horizontal distance traveled by the shot. Since the horizontal velocity remains constant, we can use the equation:
distance = velocity * time
distance = Vx * t
≈ 15.6 m/s * 1.6 s
≈ 24.96 m
Therefore, the horizontal distance traveled by the shot is approximately 24.96 meters.