A 55 kg pole vaulter falls from rest from a height of 5m onto a foam rubber pad. The pole vaulter comes to rest .3s after landing on the pad.
A) calculate the athlete's velocity just before reaching the pad
B) calculate the constant force exerted on the pole vaulter due to the collision
A) Well, this is quite a fall for our pole vaulter! To calculate their velocity just before reaching the pad, we can use the equation:
v = u + gt
Where:
v = final velocity (which we need to find)
u = initial velocity (which is zero since they start from rest)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to fall (which is 5 meters divided by their final velocity)
So, let's rearrange the equation to solve for v:
v = gt
v = 9.8 m/s^2 * 0.3 s
v ≈ 2.94 m/s
So, the athlete's velocity just before reaching the pad is approximately 2.94 m/s.
B) Now, let's calculate the constant force exerted on the pole vaulter due to the collision. To do this, we'll use Newton's second law of motion:
F = m * a
Where:
F = force (which we need to find)
m = mass of the pole vaulter (55 kg)
a = acceleration (which we can find using the equation a = Δv / t)
Since the pole vaulter comes to rest, their final velocity is zero. So, we have:
a = Δv / t
a = (0 - 2.94 m/s) / 0.3 s
a ≈ -9.8 m/s^2
So, the acceleration (deceleration, actually) is approximately -9.8 m/s^2. Now, we can calculate the force:
F = 55 kg * -9.8 m/s^2
F ≈ -539 N
Now, that's a negative force, which means it's directed upwards. In other words, it's pushing against gravity.
So, the constant force exerted on the pole vaulter due to the collision is approximately 539 N, and it's determined by their mass and the deceleration.
A) To calculate the athlete's velocity just before reaching the pad, we can use the equation of motion:
v = u + at
where:
v = final velocity (which is 0 m/s since the athlete comes to rest)
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity is what we need to find. The acceleration can be calculated using the equation:
a = (v - u) / t
Since the final velocity (v) is 0 m/s, the equation becomes:
a = -u / t
Rearranging the equation, we have:
u = -a * t
Given:
a = acceleration = g (acceleration due to gravity) = 9.8 m/s^2 (assuming no other forces acting on the athlete during the fall)
t = time = 0.3 s
Substituting the values into the equation, we get:
u = -9.8 m/s^2 * 0.3 s = -2.94 m/s
Therefore, the athlete's velocity just before reaching the pad is approximately -2.94 m/s (downwards).
B) To calculate the constant force exerted on the pole vaulter due to the collision, we can use the impulse-momentum equation:
F * t = m * (v - u)
where:
F = force
t = time
m = mass
v = final velocity (which is 0 m/s since the athlete comes to rest)
u = initial velocity
Simplifying the equation, we have:
F = (m * (v - u)) / t
Given:
m = mass = 55 kg
v = final velocity = 0 m/s
u = initial velocity = -2.94 m/s (negative sign indicates downward motion)
t = time = 0.3 s
Substituting the values into the equation, we get:
F = (55 kg * (0 - (-2.94 m/s))) / 0.3 s
F = (55 kg * 2.94 m/s) / 0.3 s
F = 540.9 N
Therefore, the constant force exerted on the pole vaulter due to the collision is approximately 540.9 Newtons.
To answer these questions, we need to use the principles of physics, specifically the laws of motion and the concept of work and energy.
A) To calculate the athlete's velocity just before reaching the pad, we can use the equation of motion. We will assume that there is no air resistance present.
The equation we need is:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s since the athlete comes to rest)
u = initial velocity (which is unknown, since the athlete starts from rest)
a = acceleration (which is due to gravity and equal to 9.8 m/s^2)
s = displacement (which is -5 m since the athlete falls downward)
Rearranging the equation, we have:
0^2 = u^2 + 2(-9.8)(-5)
Simplifying this, we get:
0 = u^2 - 98
Rearranging again, we have:
u^2 = 98
Taking the square root of both sides, we find:
u = √98 ≈ 9.90 m/s
Therefore, the athlete's velocity just before reaching the pad was approximately 9.90 m/s.
B) To calculate the constant force exerted on the pole vaulter due to the collision, we can use the work-energy principle.
The equation for work is given by:
Work = Force x Distance
Using this, we can calculate the work done by the force exerted during the collision.
The work done is equal to the change in kinetic energy (KE) of the pole vaulter. Since the athlete comes to rest, the change in KE is equal to the initial KE.
The equation for KE is given by:
KE = 0.5mv^2
Where:
m = mass of the pole vaulter (55 kg)
v = velocity just before reaching the pad (9.90 m/s)
Plugging in the values, we get:
KE = 0.5 * 55 * (9.90)^2
Calculating this, we find:
KE ≈ 2714.95 J
Since the work done is equal to the change in KE, the work done is also equal to 2714.95 J.
Therefore, the constant force exerted on the pole vaulter due to the collision is approximately 2714.95 N.
same as the last problem but you need initial velocity to get initial momentum
v = Vi - g t
v = 0 - g t so t = v/g
h = Hi + Vi t - (1/2) g t^2
0 = 5 + 0 t - (g/2) t^2
0 = 5 - (g/2)(v^2/g^2)
10 = v^2/g
v^2 = 10 g = 98
or use energy
(1/2) m v^2 = m g h
v^2 = 2 g h = 10 g = 98