A 44.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 44.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.44 102 m/s from the top of a cliff 130 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?
1. ?J

(b) Suppose the projectile is traveling 102.1 m/s at its maximum height of y = 346 m. How much work has been done on the projectile by air friction?
2. ?J

(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?
3 ?m/s

Explanation / Answer

mass m = 44 kg initial speed u = 144 m / s height of the cliff h = 130 m (a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2          E = 56056 + 456192              = 512248 J (b). maximum height H = 346 m speed at maximum height v = 102.1 m / s total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2           E ' = 149195.2 + 229337.02                = 378532.22 J therefore work done by air friction W = E' - E = -133715.78 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             = 378532.22 J - 200573.67 J             = 177958.55 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]                                           = 89.93 m / s mass m = 44 kg initial speed u = 144 m / s height of the cliff h = 130 m (a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2          E = 56056 + 456192              = 512248 J (b). maximum height H = 346 m speed at maximum height v = 102.1 m / s total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2           E ' = 149195.2 + 229337.02                = 378532.22 J therefore work done by air friction W = E' - E = -133715.78 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             = 378532.22 J - 200573.67 J             = 177958.55 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]                                           = 89.93 m / s

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