A hang glider moving at speed 9.0 m/s dives to an altitude 4.2 m lower. Ignoring drag, how fast is it then moving?
9.8*4.2=41.16
41.16/9=answer
To calculate the speed of the hang glider after diving to a lower altitude, we can use the principle of conservation of energy. Since we are ignoring drag, we can assume no energy is lost during the dive.
The conservation of energy states that the total mechanical energy of a system remains constant if no external work is done on it. In this case, the mechanical energy of the hang glider is conserved.
The mechanical energy of the hang glider can be defined as the sum of its kinetic energy (KE) and its potential energy (PE):
Total Mechanical Energy (E) = KE + PE
Before the dive:
The initial kinetic energy (KE1) is given by:
KE1 = 0.5 * m * v1^2
where m is the mass of the hang glider and v1 is its initial velocity, given as 9.0 m/s.
The initial potential energy (PE1) is given by:
PE1 = m * g * h1
where m is the mass of the hang glider, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h1 is the initial altitude.
During the dive:
The final potential energy (PE2) is given by:
PE2 = m * g * h2
where h2 is the final altitude, which is 4.2 m lower than the initial altitude.
Since the mechanical energy E is conserved, we can write the equation:
KE1 + PE1 = KE2 + PE2
Since no energy is lost due to drag, the only change in the energy is the change in altitude, resulting in a change in potential energy. Therefore, the final kinetic energy (KE2) remains the same as KE1, and we can simplify the equation to:
PE1 = PE2
Substituting the expressions for PE1 and PE2, we have:
m * g * h1 = m * g * h2
The mass m and acceleration due to gravity g cancel out, giving us:
h1 = h2
This means that the final altitude h2 is the same as the initial altitude h1.
Therefore, the speed of the hang glider after diving to a lower altitude remains the same as its initial speed, which is 9.0 m/s.