Question
I just cannot get c....
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.38 102 m/s from the top of a cliff 164 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?
answer: 600976.8
(b) Suppose the projectile is traveling 97.8 m/s at its maximum height of y = 362 m. How much work has been done on the projectile by air friction?
answer: 151155.72
(c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
Explanation / Answer
mass m = 54 kg initial speed u = 138 m / s height of the cliff h = 164 m (a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2 E = 86788.8 + 514188 = 600976.8 J (b). maximum height H = 362 m speed at maximum height v = 97.8 m / s total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2 E ' = 191570.4+ 258250.68 = 449821.08 J therefore work done by air friction W = E' - E = -151155.72 J here negative sign indicates the work is done aginst the air friction. (c). total energy when it hits the ground E " = E ' - work done by air friction in down ward motion E " = E ' - 1.5 W = 449821.08 J - 226733.58 J = 223087.5 J E " is in the form of kinetic energy .Since PE at ground is zero So, ( 1/ 2) mV^ 2 = E " from this required speed V = [ 2E" / m] = 90.89 m / s mass m = 54 kg initial speed u = 138 m / s height of the cliff h = 164 m (a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2 E = 86788.8 + 514188 = 600976.8 J (b). maximum height H = 362 m speed at maximum height v = 97.8 m / s total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2 E ' = 191570.4+ 258250.68 = 449821.08 J therefore work done by air friction W = E' - E = -151155.72 J here negative sign indicates the work is done aginst the air friction. (c). total energy when it hits the ground E " = E ' - work done by air friction in down ward motion E " = E ' - 1.5 W = 449821.08 J - 226733.58 J = 223087.5 J E " is in the form of kinetic energy .Since PE at ground is zero So, ( 1/ 2) mV^ 2 = E " from this required speed V = [ 2E" / m] = 90.89 m / s mass m = 54 kg initial speed u = 138 m / s height of the cliff h = 164 m (a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2 E = 86788.8 + 514188 = 600976.8 J (b). maximum height H = 362 m speed at maximum height v = 97.8 m / s total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2 E ' = 191570.4+ 258250.68 = 449821.08 J therefore work done by air friction W = E' - E = -151155.72 J here negative sign indicates the work is done aginst the air friction. (c). total energy when it hits the ground E " = E ' - work done by air friction in down ward motion E " = E ' - 1.5 W = 449821.08 J - 226733.58 J = 223087.5 J E " is in the form of kinetic energy .Since PE at ground is zero So, ( 1/ 2) mV^ 2 = E " from this required speed V = [ 2E" / m] = 90.89 m / s