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A shot-putter throws the shot with an initial speed of 15.5 m/s at 34.0° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground. a. 37.9 meters

b. 41.2 meters
c. 45.6 meters
d. 49.8 meters

User Michelpm
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1 Answer

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Final answer:

The horizontal distance traveled by the shot is 22.58 meters.

Step-by-step explanation:

To calculate the horizontal distance traveled by the shot, we will use the horizontal and vertical components of the initial velocity, as well as the time of flight.

The horizontal component of the initial velocity is given by vx = v * cos(θ), where v is the initial speed and θ is the angle. Plugging in the values, we get vx = 15.5 m/s * cos(34.0°) = 12.90 m/s.

The vertical component of the initial velocity is given by vy = v * sin(θ), where v is the initial speed and θ is the angle. Plugging in the values, we get vy = 15.5 m/s * sin(34.0°) = 8.62 m/s.

The time of flight can be calculated using the vertical equation of motion: Δy = vy * t + (1/2) * g * t^2, where Δy is the vertical displacement, vy is the vertical component of velocity, g is the acceleration due to gravity, and t is the time of flight. Since the shot starts and ends at the same height, Δy = 0, and the equation becomes 0 = vy * t + (1/2) * g * t^2. Solving for t, we get t = (2 * vy) / g = (2 * 8.62 m/s) / 9.8 m/s^2 = 1.75 s.

Finally, the horizontal distance traveled by the shot is given by d = vx * t = 12.90 m/s * 1.75 s = 22.58 m.

User Miboper
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