Question

A shot-putter throws the "shot" (mass = 7.3 kg ) with an initial speed of 14.9 m/s at a 40.0 ∘ angle to the horizontal. calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00 m above the ground.

Answer 1

The horizontal distance traveled by the shot is 22.4 meters.

Step-by-step explanation:

To calculate the horizontal distance traveled by the shot, we need to find the time of flight and then use it to calculate the horizontal distance. Since the shot was thrown at an angle, we can use the vertical component of the initial velocity to find the time of flight. The vertical component can be found using the equation v_y = v_initial * sin(theta), where v_initial is the initial velocity and theta is the launch angle.

Using the given values, we can calculate the vertical component of the initial velocity: v_y = 14.9 m/s * sin(40.0°) = 9.59 m/s

Next, we can use the vertical component of the initial velocity and the acceleration due to gravity to find the time of flight: t = (2 * v_y) / g = (2 * 9.59 m/s) / 9.8 m/s^2 = 1.96 s

Finally, we can use the time of flight and the horizontal component of the initial velocity to find the horizontal distance traveled by the shot. The horizontal component can be found using the equation v_x = v_initial * cos(theta).

Using the given values, we can calculate the horizontal component of the initial velocity: v_x = 14.9 m/s * cos(40.0°) = 11.4 m/s

Now we can calculate the horizontal distance traveled by the shot: d = v_x * t = 11.4 m/s * 1.96 s = 22.4 m.