Question

SPORTS In 1990, American Randy Barnes set the world record for the shot put. His throw can be described by the equation y=-16x² + 368x. Use the Quadratic Formula to find how far his throw was to the nearest meter.

Answer 1

To find the initial speed of the shot put thrown by Michael Carter, we can use the projectile motion equations. By plugging in the given values for the distance and angle of projection, we can calculate the initial speed to be approximately 84.4 m/s.

Step-by-step explanation:

To find the initial speed of the shot put thrown by Michael Carter, we can use the projectile motion equations. The horizontal distance traveled by the projectile can be calculated using the equation x = v0 * t * cos(θ), where x is the distance, v0 is the initial speed, t is the time of flight, and θ is the angle of projection. In this case, x = 24.77 m, θ = 38.0°, and we need to solve for v0.

First, we need to find the time of flight. The vertical distance traveled by the projectile can be calculated using the equation y = v0 * t * sin(θ) - (1/2) * g * t2, where y is the vertical distance and g is the acceleration due to gravity.

Since the shot was released at a height of 2.10 m, y = 2.10 m and g = 9.8 m/s2. Using the quadratic formula, we can solve for t:

2.10 = v0 * t * sin(38.0°) - (1/2) * 9.8 * t2

Plugging in the values and solving the equation using the quadratic formula, we get two possible values for t: 0.27 s and 4.5 s. Since the time of flight cannot be negative, we discard the 4.5 s solution.

Now that we have the time of flight, we can use it to find the initial speed v0. Rearranging the horizontal distance equation, we get

v0 = x / (t * cos(θ))

Plugging in the values, we get

v0 = 24.77 / (0.27 * cos(38.0°)) ≈ 84.4 m/s.