At the moment when a shot putter releases a 5.00 kg shot, the shot is 2.00 m above the ground and travelling at 15.0 m/s. It reaches a maximum height of 8.0 m above the ground and then falls to the ground. Assume air resistance is negligible.
a) What was the velocity of the shot at its maximum height?
I get this and can you check my answer of 10.36 m/s.
b) What was the velocity of the shot just before it strikes the ground?
I don't get how to do this.
c) As the shot comes to rest on the ground it creates a divet of 3.0 cm. What is the retarding force of the ground?
I also don't get this.
a) To find the velocity of the shot at its maximum height, you can use the principle of conservation of energy. At the highest point of its trajectory, the shot has potential energy equal to its initial kinetic energy.
Let's calculate the initial kinetic energy (KE_initial) and potential energy at maximum height (PE_max):
Initial kinetic energy: KE_initial = (1/2) * mass * velocity^2
KE_initial = (1/2) * 5.00 kg * (15.0 m/s)^2
KE_initial = 562.5 J
Potential energy at maximum height: PE_max = mass * gravity * height
PE_max = 5.00 kg * 9.8 m/s^2 * 8.0 m
PE_max = 392 J
Since there is no significant loss of energy due to air resistance, we can equate these two energy forms:
KE_initial = PE_max
562.5 J = 392 J
Now, we can solve for the velocity at the maximum height (v_max) using the kinetic energy equation:
KE = (1/2) * mass * velocity^2
392 J = (1/2) * 5.00 kg * v_max^2
Rearranging the equation and solving for v_max:
v_max^2 = (392 J * 2) / 5.00 kg
v_max^2 = 156.8 J / 5.00 kg
v_max^2 = 31.36 m^2/s^2
Taking the square root of both sides, we get:
v_max = √(31.36 m^2/s^2)
v_max ≈ 5.60 m/s
Hence, your answer of 10.36 m/s for part a) is incorrect. The correct velocity of the shot at its maximum height is approximately 5.60 m/s.
b) To determine the velocity of the shot just before it strikes the ground, we can use the principle of conservation of mechanical energy (assuming again there is no significant energy loss due to air resistance). At the highest point, the kinetic energy is fully converted into potential energy. Therefore, as the shot approaches the ground, the potential energy will be converted back into kinetic energy.
The potential energy at the maximum height (PE_max) is the same as the potential energy just before striking the ground (PE_final). To find the final kinetic energy (KE_final) and thus the final velocity (v_final), we can set PE_max = KE_final and solve for v_final.
PE_max = KE_final
392 J = (1/2) * 5.00 kg * v_final^2
Rearranging the equation and solving for v_final:
v_final^2 = (392 J * 2) / 5.00 kg
v_final^2 = 156.8 J / 5.00 kg
v_final^2 = 31.36 m^2/s^2
Taking the square root of both sides, we get:
v_final = √(31.36 m^2/s^2)
v_final ≈ 5.60 m/s
Hence, the velocity of the shot just before it strikes the ground is approximately 5.60 m/s.
c) To determine the retarding force of the ground, we need to calculate the deceleration (negative acceleration) experienced by the shot as it comes to rest on the ground. We can use the equation:
Final velocity^2 = Initial velocity^2 + 2 * acceleration * distance
Let's assume the final velocity (v_f) is 0 m/s, the initial velocity (v_i) is the velocity just before it strikes the ground (which is approximately 5.60 m/s), and the distance (d) is the depth of the divet created (which is 0.03 m or 3.0 cm):
0 m/s = (5.60 m/s)^2 + 2 * acceleration * 0.03 m
Rearranging the equation and solving for the acceleration:
2 * acceleration * 0.03 m = -(5.60 m/s)^2
acceleration = -((5.60 m/s)^2) / 0.06 m
acceleration ≈ -330.67 m/s^2
The retarding force (or negative force) exerted by the ground is given by Newton's second law:
Force = mass * acceleration
Assuming the mass (m) is still 5.00 kg, we can calculate the retarding force:
Force = 5.00 kg * -330.67 m/s^2
Force ≈ -1653.35 N (rounded to two decimal places)
Therefore, the retarding force of the ground is approximately -1653.35 N (negative sign indicates that it acts opposite to the motion).
a) To find the velocity of the shot at its maximum height, you can use the principle of conservation of mechanical energy. At the maximum height, the shot has reached its highest potential energy, meaning all its initial kinetic energy has been converted into potential energy. You can set the initial kinetic energy equal to the potential energy at the maximum height:
1/2mv^2 = mgh
Here, m is the mass of the shot (5.00 kg), v is the velocity of the shot, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground (8.0 m).
Plugging in the values:
1/2(5.00 kg)(v^2) = (5.00 kg)(9.8 m/s^2)(8.0 m)
simplifying:
2.5v^2 = 392
dividing both sides by 2.5:
v^2 = 156.8
taking the square root of both sides:
v ≈ 12.52 m/s
So, the velocity of the shot at its maximum height is approximately 12.52 m/s. It seems that your previous answer of 10.36 m/s is incorrect.
b) To find the velocity of the shot just before it strikes the ground, you can use the conservation of mechanical energy again. At the maximum height, the shot has no potential energy left and has converted all of it back into kinetic energy. The velocity just before it strikes the ground can be found using:
1/2mv^2 = mgh
Here, m is the mass of the shot (5.00 kg), v is the velocity of the shot, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground (0 m).
Plugging in the values:
1/2(5.00 kg)(v^2) = (5.00 kg)(9.8 m/s^2)(0 m)
Simplifying:
2.5v^2 = 0
Since 2.5v^2 is zero, it means that v must also be zero. Therefore, the velocity of the shot just before it strikes the ground is 0 m/s.
c) To find the retarding force of the ground, you first need to determine the impulse experienced by the shot when it strikes the ground. Impulse is equal to the change in momentum. The initial momentum of the shot can be calculated using:
initial momentum = mass × initial velocity
Here, the mass of the shot is 5.00 kg and the initial velocity is 15.0 m/s. Therefore:
initial momentum = 5.00 kg × 15.0 m/s = 75 kg·m/s
When the shot comes to rest on the ground, its final momentum is 0 kg·m/s. The change in momentum is then:
change in momentum = final momentum - initial momentum
= 0 kg·m/s - 75 kg·m/s
= -75 kg·m/s
The impulse experienced by the shot can be given by the change in momentum:
impulse = force × time
Since the impulse is calculated using the average force and the time of impact, the formula can be written as:
impulse = average force × time
Assuming the time of impact is very short, we can approximate the average force as the retarding force of the ground. Rearranging the formula, we can solve for the retarding force:
average force = impulse / time
Plugging in the values:
average force = -75 kg·m/s / (time)
The value of time is not given in the question, so it is not possible to calculate the retarding force of the ground without that information.
goes up from 2 to 8 or 6 meters
do vertical problem
v = Vi - g t
0 = Vi - 9.8 t at the top (8meters)
so
t at top = Vi/9.8
h = 2 + Vi t - 4.9 t^2
at top
8 = 2 + Vi t -(1/2)(9.8) t^2
6 = Vi t -(1/2)(9.8) t^2
6 = Vi^2/9.8 - (1/2) Vi^2/9.8
12 (9.8) = Vi^2
Vi = 10.84 m/s
now do the combined vertical and horizontal
initial
sin A = 10.84/15
so A = angle up from horizontal = 46.3 degrees
horizontal speed = u = 15 cos 46.3 = 10.37 m/s forever (agree with you)
b) v = Vi - 9.8 t
h = 0 at ground
0 = 2 + 10.84 t - 4.9 t^2
4.9 t^2 - 10.84 t - 2 = 0
t = [ 10.84 +/- sqrt (118+39.2) ]/9.8
t = 2.39 seconds to ground
so
v = 10.84 - 9.8(2.39) = - 12.5 m/s
u is still 10.37
so speed = sqrt(12.5^2+10.^2)
tan angle below hor = 12.5/10.4
c) just vertical problem. It does not say how long the hole is
Force * time = change of momentum
F (distance/average speed) = 5(12.5)
average speed = 12.5/2
F (.03/6.25) = 5(12.5)