At the 2004 Olympic Games in Athens, Dwight
Phillips won the gold medal in men’s long
jump with a jump of 8.59 m. If the angle of
his jump was 23°, what was his takeoff speed?
(Treat the jumper as an object; ignore that his
legs are in front of him when he lands.)
range = Vo^2 sin(2Θ) / g
Vo^2 = sin(46º) / (8.59 m * 9.81 m/s^2)
oops... upside down
Vo^2 = 8.59 m * 9.81 m/s^2 / sin(46º)
To determine Dwight Phillips' takeoff speed, we can use the principle of conservation of mechanical energy.
The mechanical energy of an object can be divided into two types: kinetic energy (KE) and potential energy (PE). When Dwight Phillips jumps, he converts his potential energy to kinetic energy.
Let me explain the steps to calculate his takeoff speed:
Step 1: Calculate the potential energy (PE) at the takeoff position using the formula: PE = m * g * h
Where:
m = mass of Dwight Phillips (which we'll assume as 1 kg for simplicity)
g = acceleration due to gravity (9.8 m/s^2)
h = height from ground level to takeoff position (which we'll consider zero for this problem, as we're only interested in speed)
Since the height is zero, the potential energy at takeoff is also zero: PE = 0.
Step 2: Calculate the kinetic energy (KE) at the takeoff position using the formula: KE = (1/2) * m * v^2
Where:
m = mass of Dwight Phillips (1 kg)
v = takeoff velocity
We want to find the takeoff velocity (v), so we rearrange the formula:
v = sqrt((2 * KE) / m)
Step 3: Equate potential energy (PE) and kinetic energy (KE) to find the takeoff speed (v):
PE = KE
Since PE is zero, we have:
0 = (1/2) * m * v^2
Step 4: Solve for v:
v^2 = 0
v = 0 m/s
So, based on these calculations, the takeoff speed for Dwight Phillips is found to be 0 m/s.