An object of mass m 2 kilograms talls vertically to the ground under the action

Question

An object of mass m 2 kilograms talls vertically to the ground under the action of the earth gravtational acceleration of magntude g that tho torco on tho objact in this situation is mg. whore th nagativo sign says the torce points downwards. 10 meters per second squared. Denote by u vertical coordinate, positive upwards, and let y 0 be at the earth surface. Recall (a) VWnte the differential equation satistied by this system. Note: Write or , write y for y), and yp for y'() (b) Find the mechanical energy Eof this system E(I) = Note: Write tor t, write y for y(t) and yp for y'(t). (c) It the initial position ot the obiect is (0) 5 meters and its initial velocity is y'(0) meters per second, find the vakue of the speed of the object l'. when the object reaches the ground.

Explanation / Answer

y'' = acceleration of body in y direction

=> y''= -g

=> y'' = -10 metre per second squared

E(t) = potential energy + kinetic energy = mgy + 1/2 mv2 = mgy(t) + 1/2m(y'(t))2

E(t) = 2*10*y(t) + 1/2 *2 *(y'(t))2 => E(t) = 20y(t)+(y'(t))2

y(0) = 5

E(0) = 20*5+0 = 100

E(t_ground) = 20*0 + (y')2 =  (y')2

Using conservation of mechanical enegry we get

=> 100 =  (y')2 => |y'| = 10 m

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