A 59 kg pole-vaulter falls from rest from a height of 5.5 m onto a foam rubber pad. The pole-vaulter comes to rest 0.39 s 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
a. mgh=1/2 mv^2 or v= sqrt2gh
b. force*time= mass*v
solve for force.
To solve this problem, we can use the equations of motion to find the velocity of the pole-vaulter just before reaching the pad. The equation we'll use is the one for finding final velocity given initial velocity, acceleration, and time:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 since the pole-vaulter falls from rest)
a = acceleration (which we need to calculate using the height and time)
t = time
Step 1: Calculate the acceleration
The acceleration can be calculated using the equation for kinematic motion:
s = ut + (1/2)at^2
where:
s = distance (height in this case)
u = initial velocity (0 m/s)
t = time (0.39 s)
a = acceleration (which we are trying to find)
Let's rearrange the equation to solve for acceleration:
s = (1/2)at^2
2s = at^2
a = (2s) / (t^2)
Given:
s = 5.5 m
t = 0.39 s
a = (2 * 5.5) / (0.39^2)
a = 72.58065 m/s^2 (rounded to 3 decimal places)
Step 2: Calculate the final velocity
Now that we have the acceleration, we can substitute it into the equation of motion to find the final velocity:
v = u + at
v = 0 + (72.58065)(0.39)
v = 28.24419 m/s (rounded to 2 decimal places)
Therefore:
a. The athlete's velocity just before reaching the pad is approximately 28.24 m/s.
b. To calculate the constant force exerted on the pole-vaulter due to the foam rubber pad, we need to use Newton's second law of motion:
F = ma
where:
F = force
m = mass (59 kg)
a = acceleration (72.58065 m/s^2)
F = (59)(72.58065)
F = 4280.64435 N (rounded to 2 decimal places)
Therefore, the constant force exerted on the pole-vaulter due to the foam rubber pad is approximately 4280.64 N.
To calculate the athlete's velocity just before reaching the pad, we can use the equation:
v = u + at
where:
v = final velocity (unknown)
u = initial velocity (0 m/s, since the athlete falls from rest)
a = acceleration (unknown)
t = time (0.39 s)
First, let's find the acceleration using the equation:
v = u + at
Since the athlete comes to rest, the final velocity (v) is 0 m/s. Therefore, our equation becomes:
0 = 0 + a * 0.39
Simplifying the equation gives:
0 = 0.39a
Dividing both sides by 0.39, we find that:
a = 0 m/s²
Now, we can use this acceleration value in the original equation:
v = u + at
v = 0 + 0 * 0.39
v = 0
Therefore, the athlete's velocity just before reaching the pad is 0 m/s.
For part b, to calculate the constant force exerted on the pole-vaulter due to the foam rubber pad, we can use Newton's second law of motion, which states:
F = m * a
where:
F = force (unknown)
m = mass of the pole-vaulter (59 kg)
a = acceleration (0 m/s², since the athlete comes to rest)
Substituting the given values into the equation:
F = 59 kg * 0 m/s²
F = 0 N
Therefore, the constant force exerted on the pole-vaulter due to the foam rubber pad is 0 N.