A 70-kg circus performer is fired from a cannon that is elevated at an angle of 43° above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 3.7 m from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 4.8 s to travel the horizontal distance of 59 m between this point and the net.
What is the launch speed? (Ignore friction and air resistance.)
What is the effective spring constant of the firing mechanism?
To solve for the launch speed, we can use the equations of motion for projectile motion. Let's break down the different components involved.
Given information:
Mass of the performer (m) = 70 kg
Angle of elevation (θ) = 43°
Vertical displacement (h) = 3.7 m
Horizontal distance (d) = 59 m
Time of flight (t) = 4.8 s
To calculate the launch speed (Vx), we will consider the vertical and horizontal components separately.
Step 1: Calculate the initial vertical velocity (Vy) using the formula:
Vy = h / t
Vy = 3.7 m / 4.8 s
Step 2: Calculate the vertical component of the launch speed (Vy) using the equation of motion:
Vy = V0 * sin(θ)
V0 = Vy / sin(θ)
Step 3: Calculate the initial horizontal velocity (Vx) using the formula:
Vx = d / t
Step 4: Calculate the launch speed (V) using the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
Now let's calculate each step:
Step 1:
Vy = 3.7 m / 4.8 s ≈ 0.771 m/s
Step 2:
V0 = Vy / sin(43°)
Step 3:
Vx = 59 m / 4.8 s
Step 4:
V = √(Vx^2 + Vy^2)
Once we calculate the values in steps 2, 3, and 4, we will have the launch speed (V) of the performer.
To find the effective spring constant of the firing mechanism, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Given information:
Mass of the performer (m) = 70 kg
Vertical displacement of the band (x) = 3.7 m
We can use the equation:
Force (F) = k * x
Step 5: Calculate the force exerted by the band (F) using the performer's weight:
F = m * g
Step 6: Calculate the effective spring constant (k) using Hooke's law:
k = F / x
By substituting the appropriate values into the equation in step 6, we will find the effective spring constant (k).
Please note that the value of g (acceleration due to gravity) is approximately 9.8 m/s².
To find the launch speed, we need to use the kinematic equations to analyze the motion of the circus performer after he becomes free of the elastic bands.
1. First, let's find the vertical component of the launch speed. Since the height above the floor is the same as that of the net, we can equate the initial vertical velocity with zero (because the performer will eventually reach the maximum height and then fall down). We can use the equation:
vf = vi + gt
where vf is the final velocity (zero in this case because the performer reaches maximum height), vi is the initial vertical velocity (which we want to find), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight (4.8 s).
Using the equation, we can rearrange it to solve for the initial vertical velocity:
vi = -gt
Substituting the values:
vi = -(9.8 m/s^2)(4.8 s)
vi = -47.04 m/s
Since the velocity is downward, we take the negative sign.
2. Next, let's find the horizontal component of the launch speed. We can use the equation:
s = vi*t
where s is the horizontal distance (59 m) and vi is the initial horizontal velocity (which we want to find), and t is the time of flight (4.8 s).
Rearranging the equation:
vi = s/t
Substituting the values:
vi = 59 m / 4.8 s
vi ≈ 12.29 m/s
3. Finally, let's find the launch speed. Since the launch speed is the resultant velocity of both the horizontal and vertical components, we can use the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
where v is the launch speed we want to find, vx is the horizontal component (12.29 m/s), and vy is the vertical component (-47.04 m/s).
Calculating:
v = sqrt((12.29 m/s)^2 + (-47.04 m/s)^2)
v ≈ 49.11 m/s
Therefore, the launch speed of the circus performer is approximately 49.11 m/s.
To find the effective spring constant of the firing mechanism, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:
F = -kx
where F is the force, k is the spring constant (which we want to find), and x is the displacement from the equilibrium position (3.7 m).
Rearranging the equation:
k = -F / x
Now, the force can be calculated using Newton's second law:
F = ma
where m is the mass of the performer (70 kg) and a is the acceleration (which is the same as the acceleration due to gravity, approximately 9.8 m/s^2).
Substituting the values:
F = (70 kg)(9.8 m/s^2)
F = 686 N
Now, substituting F and x into the spring constant equation:
k = -(686 N) / (3.7 m)
k ≈ -185.68 N/m
Thus, the effective spring constant of the firing mechanism is approximately -185.68 N/m. Note that the negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.
Dx = Xo * T = 59 m.
Xo * 4.8 = 59
Xo = 12.29 m/s = Hor. component of
initial velocity.
Vo = Xo/cosA = 12.29/cos43 = 16.81 m/s
k = F/d = m*g/d = 72*9.8/3.7m=191N/m