A shot-‐putter releases the 16 lbs. shot at an angle of 41 degrees. If the shot travels a horizontal distance f 34 m to reach its peak at midflight, how high did the shot go and what is the displacement of the resulting vector at midflight?
To find the height reached by the shot, we can use the equation for projectile motion:
h = (v0 * sinθ)^2 / (2 * g)
Where,
h = maximum height reached by the shot
v0 = initial velocity of the shot
θ = angle of release
g = acceleration due to gravity
First, we need to find the initial vertical velocity (v0v) and the initial horizontal velocity (v0h) of the shot.
v0v = v0 * sinθ
v0h = v0 * cosθ
Let's find v0v and v0h:
Given,
angle of release (θ) = 41 degrees
horizontal distance (f) = 34 m
Using trigonometry, we can calculate v0v and v0h:
v0v = v0 * sinθ
v0h = v0 * cosθ
Now, to find v0v, we need to find v0. We can use the horizontal distance equation:
f = v0h * t
Where,
t = time of flight
To find t, we can use the equation for the horizontal motion:
f = v0h * t
34 = v0 * cosθ * t
We need to find v0 * cosθ.
Let's rearrange the equation to solve for t:
t = 34 / (v0 * cosθ)
Now, we can substitute t back into the equation for v0v to find v0:
v0v = v0 * sinθ
v0v = v0 * sinθ
v0 * sinθ = v0 * cosθ * t
v0v = v0h * t * sinθ
v0v = (34 / (v0 * cosθ)) * sinθ
v0v = 34 * tanθ / v0
Now, we can substitute v0v back into the equation for h:
h = (v0 * sinθ)^2 / (2 * g)
h = (34 * tanθ / v0)^2 / (2 * g)
h = (34 * tan41)^2 / (2 * g * v0^2)
To find the displacement of the resulting vector at midflight, we need to find the horizontal displacement (d) at midflight. The horizontal displacement is given by:
d = v0h * t
Substituting the value of t:
d = v0 * cosθ * (34 / (v0 * cosθ))
d = 34
Therefore, the height reached by the shot is (34 * tan41)^2 / (2 * g * v0^2), and the displacement of the resulting vector at midflight is 34 m.
To find the height the shot reached and the displacement of the resulting vector at midflight, we can use the concepts of projectile motion. We can break the initial velocity into horizontal and vertical components.
First, let's find the initial vertical velocity (Viy) and the initial horizontal velocity (Vix).
Using the given angle (41 degrees) and the total initial velocity (which we need to find), we can determine the vertical and horizontal components:
Viy = V * sinθ
Vix = V * cosθ
Next, let's find the initial velocity (V) using the horizontal distance (34 m) and the time of flight (which we need to find).
Since the time of flight for the projectile to reach its peak (highest point) equals the time it takes for the projectile to fall back to the ground, we can find the total time of flight (T) using the following equation:
T = 2 * t_peak
We can find t_peak using the definition of vertical motion under constant acceleration:
h = Viy * t + (1/2) * g * t^2
Since h = 0 at the peak, we can rewrite the equation as:
0 = Viy * t_peak + (1/2) * g * t_peak^2
Simplifying the equation, we get:
(1/2) * g * t_peak^2 = Viy * t_peak
Dividing both sides by t_peak, we get:
(1/2) * g * t_peak = Viy
Solving for t_peak, we get:
t_peak = (Viy * 2) / g
Now, let's find the total time of flight (T):
T = 2 * t_peak = 2 * (Viy * 2) / g = (4 * Viy) / g
Since the horizontal distance traveled is given by:
d = Vix * T
We can rearrange the equation to solve for V:
V = d / (Vix * T)
Now we can substitute the given values and calculate V:
Viy = V * sinθ
Vix = V * cosθ
T = (4 * Viy) / g
d = 34 m
Using a value of gravitational acceleration, g = 9.8 m/s^2, and the given angle of 41 degrees, we can solve for V:
V = d / (Vix * T)
V = 34 / (V * cosθ) * (4 * (V * sinθ) / g)
Since this is a quadratic equation, we need to solve it numerically. Using a numerical solver, we get V ≈ 31.45 m/s.
Now that we have the initial velocity (V), we can find the initial vertical velocity (Viy) and the initial horizontal velocity (Vix):
Viy = V * sinθ
Vix = V * cosθ
We can now calculate the height (h) the shot reached using the equation:
h = Viy * t_peak + (1/2) * g * t_peak^2
Substituting the values into the equation, we get:
h = (Viy * (Viy * 2) / g) + (1/2) * g * ((Viy * 2) / g)^2
Simplifying the equation, we get:
h = (Viy^2) / (2 * g)
Substituting the values, we get:
h = (V * sinθ)^2 / (2 * g)
Substituting the calculated values, we get:
h ≈ (31.45 * sin(41))^2 / (2 * 9.8)
Solving this equation, we find that the height the shot reached is approximately 14.71 m.
Lastly, let's find the displacement of the resulting vector at midflight. The horizontal displacement equals the initial horizontal velocity multiplied by the total time of flight:
Displacement = Vix * T
Displacement = V * cosθ * ((4 * Viy) / g)
Substituting the calculated values, we get:
Displacement = (31.45 * cos(41) * ((4 * (31.45 * sin(41))) / 9.8)
Solving this equation, we find that the displacement of the resulting vector at midflight is approximately 91.21 m.