A shot-putter throws the shot ( mass = 7.3 kg) with an initial speed of 14.0 m/s at a 36.0 ∘ 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.
To calculate the horizontal distance traveled by the shot, we can use the horizontal component of the initial velocity and the time of flight of the shot.
First, let's find the initial horizontal velocity of the shot. We can use trigonometry to determine this. The horizontal component of the initial velocity can be found by multiplying the initial speed by the cosine of the launch angle:
Horizontal velocity (Vx) = initial speed × cos(angle)
Vx = 14.0 m/s × cos(36.0°)
Vx ≈ 11.34 m/s
Now, let's calculate the time of flight for the shot. We can use the vertical component of the initial velocity and the acceleration due to gravity to determine this. The vertical component of the initial velocity can be found by multiplying the initial speed by the sine of the launch angle:
Vertical velocity (Vy) = initial speed × sin(angle)
Vy = 14.0 m/s × sin(36.0°)
Vy ≈ 8.05 m/s
The time of flight can be determined by dividing twice the vertical velocity component by the acceleration due to gravity (9.8 m/s²):
Time of flight = 2 × Vy / gravity
Time of flight = 2 × 8.05 m/s / 9.8 m/s²
Time of flight ≈ 1.64 seconds
Finally, we can calculate the horizontal distance traveled by multiplying the horizontal velocity by the time of flight:
Horizontal distance = Vx × time of flight
Horizontal distance = 11.34 m/s × 1.64 seconds
Horizontal distance ≈ 18.58 meters
Therefore, the horizontal distance traveled by the shot would be approximately 18.58 meters.
To calculate the horizontal distance traveled by the shot, we can use the projectile motion equations. The horizontal motion and vertical motion are independent of each other.
First, let's calculate the time it takes for the shot to reach the ground. We can do this by using the vertical motion equation:
h = v_iy * t + (1/2) * g * t^2
where:
h = initial height = 2.20 m
v_iy = initial vertical velocity = v_i * sin(theta)
g = acceleration due to gravity = 9.8 m/s^2
Substituting the given values:
2.20 m = (14.0 m/s * sin(36.0°)) * t + (1/2) * (9.8 m/s^2) * t^2
Simplifying the equation:
4.9 t^2 + 7.0 t - 2.2 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = 4.9, b = 7.0, and c = -2.2.
Using the quadratic formula, we get two possible values for t: t = 0.226 s or t = -1.652 s. Since time cannot be negative, the correct answer is t = 0.226 s.
Now that we have the time of flight, we can calculate the horizontal distance using the horizontal motion equation:
d = v_ix * t
where:
v_ix = initial horizontal velocity = v_i * cos(theta)
t = time of flight = 0.226 s
Substituting the given values:
d = (14.0 m/s * cos(36.0°)) * 0.226 s
Calculating the horizontal distance, we get:
d ≈ 2.67 m
Therefore, the shot travels approximately 2.67 meters horizontally.
the height is
y = 2.2 + 14*sin(36)t - 4.9t^2
solve for t when y=0
The horizontal speed is a constant 14*cos(36), so now you can see how far it traveled.