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 1.34 102 m/s from the top of a cliff 152 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? J (b) Suppose the projectile is traveling 95.0 m/s at its maximum height of y = 339 m. How much work has been done on the projectile by air friction? 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? m/s

Explanation / Answer

mass m = 44 kg

initial speed u = 134 m / s

height of the cliff h = 152 m

(a). the initial total mechanical energy of the projectile E = mgh + ( 1/ 2) mu^ 2
E =44*9.8* 152 + 0.5 * 44 *134^2

E= 460574.4 J

(b). maximum height H = 339m

speed at maximum height v = 95 m / s

total mechanical energy at maximum height E ' = mgH + ( 1/ 2) mv^ 2
E' = 44*9.8* 339 + 0.5*44* 95^2

E ' = 344726.8 J

therefore work done by air friction W = E' - E = -115848J

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

= 344726.8 - 173772J

= 1709548 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]

= 88.15m / s(approximately)

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