A shot-putter throws the shot (mass = 7.3kg) with an initial speed of 15.0 m/s at a 33.0 degree angle to the horizontal.
Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.00m above the ground.
no air resistance?
consider the horizonal distance:
d= Vihorizontal*time = 15cos33*t
now the vertical:
finalheight=initialheight+ vivertical*t- 4.9t^2
or
0=2+15sin33 t - 4.9 t^2
solve the second equation for time. Use the quadratic equation. Put that time t into the first equation (distance horizontal).
sdds
disco
To solve this problem, we can break down the initial velocity of the shot into its horizontal and vertical components.
The horizontal component of the initial velocity (Vx) can be found using the equation Vx = V * cosθ, where V is the magnitude of the initial velocity (15.0 m/s) and θ is the launch angle (33.0 degrees).
Vx = 15.0 m/s * cos(33.0 degrees)
Vx = 15.0 m/s * 0.8387
Vx ≈ 12.58 m/s
The vertical component of the initial velocity (Vy) can be found using the equation Vy = V * sinθ.
Vy = 15.0 m/s * sin(33.0 degrees)
Vy = 15.0 m/s * 0.5446
Vy ≈ 8.17 m/s
Now that we have the initial horizontal (Vx) and vertical (Vy) components of the velocity, we can solve for the time it takes for the shot to reach the ground.
Using the equation h = Vy * t + (1/2) * g * t^2, where h is the initial height (2.00 m), Vy is the vertical velocity (8.17 m/s), g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time.
Plugging in the values:
2.00 m = 8.17 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation:
0 = 4.9 m/s^2 * t^2 + 8.17 m/s * t - 2.00 m
We can solve this quadratic equation by using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 4.9 m/s^2, b = 8.17 m/s, and c = -2.00 m.
Using the quadratic formula:
t = (-8.17 ± √(8.17^2 - 4 * 4.9 * -2.00)) / (2 * 4.9)
t ≈ 0.592 s (ignoring the negative value)
Now that we have the time it takes for the shot to reach the ground, we can calculate the horizontal distance traveled by the shot.
Horizontal distance (d) = Vx * t
d = 12.58 m/s * 0.592 s
d ≈ 7.45 meters
Therefore, the horizontal distance traveled by the shot is approximately 7.45 meters.