A 50.7-g golf ball is driven from the tee with an initial speed of 51.2 m/s and rises to a height of 21.8 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.55 m below its highest point?
a. Y^2 = Yo^2 + 2g*h = 0 @ max. ht.
Yo^2 = -2g*h = 19.6 * 21.8 = 427.28
Yo = 20.7 m/s. = Ver. component of initial velocity.
Yo = 51.2*sin A = 20.7 m/s.
sinA = 20.7/51.2 = 0.40373
A = 23.8o.
Vo = 51.2[23.8o]
Xo = 51.2*cos23.8 = 46.8 m/s. = Hor.
component of initial velocity.
V = Xo + Yi
V = 46.8 + 0i @ max. Ht.
V = Xo = 46.8 m/s @ max. Ht.
KE=0.5m*V^2=0.5*0.0507*(46.8)^2=55.5 J.
b. Y^2 = Yo^2 + 2g*h
Y^2 = (20.7)^2 + (-19.6*(21.8-6.55)
Y^2 = 428.49 - 298.9 = 129.59
Y = 11.38 m/s. = Ver. component of velocity.
V = Xo + Yi
V = 46.8 + 11.4i = 48.2 m/s[13.7o].
Correction:
b. Y^2 = Yo^2 + 2g*h
Y^2 = 0 + 19.8(21.8-6.55) = 299
Y = 17.3 m/s. = Ver. component.
V = Xo+Yi = 46.8 + 17.3i
V = sqrt(46.8^2+17.3^2) = 49.9 m/s.
To solve part (a) of the problem, we need to find the kinetic energy of the golf ball at its highest point.
The kinetic energy (KE) of an object can be calculated using the formula:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
In this case, we are given the mass of the golf ball, which is 50.7 g. However, we need to convert it to kilograms (kg) because the SI unit for mass is kilograms.
1 kg = 1000 g
So, the mass of the golf ball in kilograms is:
m = 50.7 g * (1 kg / 1000 g) = 0.0507 kg
Next, we are given the initial speed of the golf ball, which is 51.2 m/s. Since we need to find the kinetic energy at the highest point, we can assume the final speed is 0 m/s because the ball is momentarily at rest at its highest point.
Using the given information, we have:
m = 0.0507 kg
v = 0 m/s
Plugging these values into the formula, we can calculate the kinetic energy:
KE = (1/2) * m * v^2
= (1/2) * 0.0507 kg * (0 m/s)^2
= 0
Therefore, the kinetic energy of the ball at its highest point, neglecting air resistance, is 0 Joules.
Moving on to part (b) of the problem, we need to find the speed of the golf ball when it is 6.55 m below its highest point.
To solve this, we can use the law of conservation of energy. At its highest point, all of the initial kinetic energy will be converted to potential energy (PE). As the ball falls, the potential energy will be converted back to kinetic energy.
The potential energy (PE) of an object is given by the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
In this case, we are given the mass of the golf ball (0.0507 kg) and the height below the highest point (6.55 m). Substituting these values into the formula, we can calculate the potential energy:
PE = 0.0507 kg * 9.8 m/s^2 * 6.55 m
= 3.208 J
Since the potential energy is converted back to kinetic energy, we can equate the two:
KE = PE
(1/2) * m * v^2 = 3.208 J
Now, we can solve for the speed (v):
v^2 = 2 * (3.208 J) / (0.0507 kg)
v^2 = 126.35 m^2/s^2
Taking the square root of both sides, we find:
v = √(126.35 m^2/s^2)
v ≈ 11.24 m/s
Therefore, the speed of the golf ball when it is 6.55 m below its highest point is approximately 11.24 m/s.