Question

A 46.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 142 m/s from the top of a cliff 150 m above level ground, where the ground is taken to be

y = 0.

(a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.)
J

(b) Suppose the projectile is traveling 100.6 m/s at its maximum height of

y = 360 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

Answer 1

mass m = 48 kg

initial speed u = 142 m / s

height of the cliff h = 150 m

(a). the initial total mechanical energy of the projectile

E = m*g*h + ( 1/ 2) m*u^ 2

E =48*9.8* 150 + 0.5 * 48 *142^2

E = 70560 + 483936

E = 554496 J

(b)

maximum height H = 360 m

speed at maximum height v = 100.6 m / s

total mechanical energy at maximum height

E ' = m*g*H + ( 1/ 2) m*v^ 2

E' = 48*9.8* 360 + 0.5*48* 100.6^2

E ' = 169344 +242888.64

E ' = 412232.64 J

therefore work done by air friction

W = E' - E

W = -142263.36 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

E '' = 423293.04 - 213395.04 J

E'' = 209898 J

E " is in the form of kinetic energy .Since PE at ground is zero

So, ( 1/ 2) m*V^ 2 = E ",

from this required speed V = sqrt[ 2E" / m]

V = sqrt [ 2*209898/48]

V = 93.51 m/s