Question

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)

The initial speed of the shot is approximately:
a) 15 m/s
b) 20 m/s
c) 25 m/s
d) 30 m/s

Answer 1

To find the initial speed of the shot in the shot put, you can use the projectile motion equations. By calculating the time it takes for the shot to reach the maximum height and using the horizontal distance traveled, you can find the initial speed. in this case, the initial speed of the shot is approximately 39.36 m/s.

Step-by-step explanation:

To find the initial speed of the shot in the shot put, we can use the projectile motion equations. First, we need to find the time it takes for the shot to reach the maximum height. Using the height and the acceleration due to gravity, we can find the time:

t = sqrt((2 * height) / g) = sqrt((2 * 2.10 m) / 9.8 m/s^2) = 0.644 s

Next, we can use the horizontal distance traveled by the shot to find the initial speed. The range can be calculated using the formula:

range = initial speed * time * cos(angle)

Substituting the given values and solving for the initial speed:

24.77 m = initial speed * 0.644 s * cos(38.0°)

initial speed = 24.77 m / (0.644 s * cos(38.0°)) = 39.36 m/s

Therefore, the initial speed of the shot is approximately 39.36 m/s.