Answer
(a) Without air resistance, the speed would be 294 m/s when the glider landed on the ground.
(b) The magnitude of the force of air resistance was 2400 N on average.
Work Step by Step
(a) $v = (480~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 133.3~m/s$
We can use conservation of energy to solve this question. Let $E_2$ be the total energy on the ground and let $E_1$ be the total energy at a height of 3500 m.
$E_2 = E_1$
$PE_2 + KE_2 = PE_1 + KE_1$
$0 + \frac{1}{2}mv_2^2 = mgh + \frac{1}{2}mv_1^2$
$v_2^2 = 2gh + v_1^2$
$v_2 = \sqrt{2gh + v_1^2}$
$v_2 = \sqrt{(2)(9.80~m/s^2)(3500~m)+(133.3~m/s)^2}$
$v_2 = 294~m/s$
Without air resistance, the speed would be 294 m/s when the glider landed on the ground.
(b) $v = (210~km/h)(\frac{1000~m}{1~km})(\frac{1~h}{3600~s}) = 58.3~m/s$
We can use work and energy to solve this question. The air resistance did negative work on the glider which reduced the final speed of the glider from its theoretical value found in part (a). The magnitude of the work is equal to the difference in the theoretical kinetic energy (without air resistance) minus the actual kinetic energy when the glider landed.
$W = \frac{1}{2}(980~kg)(294~m/s)^2 - \frac{1}{2}(980~kg)(58.3~m/s)^2$
$W = 4.069 \times 10^7~J$
We need to find the distance $d$ that the glider traveled as it came in for landing.
$d = \frac{3500~m}{sin(12^{\circ})} = 16,800~m$
$F\cdot d = W$
$F = \frac{W}{d} = \frac{4.069 \times 10^7~J}{16,800~m}$
$F = 2400~N$
The magnitude of the force of air resistance was 2400 N on average.
