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Question

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A very small steel marble is shown rolling at a constant speed on a horizontal table. The marble leaves the table at P, falls, and hits the ground at M. This is illustrated in the diagram below which is drawn to scale. Calculate the time it took the marble to travel from L to M.

and

An artillery shell is launched on a flat, horizontal field at an angle of = 44.0° with respect to the horizontal and with an initial speed of v0 = 253 m/s. What is the horizontal distance covered by the shell after 5.81 s of flight?

What is the height of the shell at this moment?

Explanation / Answer

FOR ONE

Apply kinematic equation to the stell marble

y- y0+ vot + 1/2 * g t^2

final height of the marble is zero y = 0

let yo is initial height be h

0 = h + vo*t - 1/2 * g * t^2
since marble is moving with constant spee intial velocityof the marble is zero

h = 1/2 * g * t^2
t^2 = 2h / g
t = root 2h/g

the initial height of the marble is h = 0.5 m

t = root 2(0.5)/9.8 = 0.319 s

from the conservation of energy

KE-PE = 0.3 m
Vo = 0.3/0.319

      = 0.94 msec
t1 = (P-M)/Vo

    = 0.5/0.94

     = 0.532 sec
t = t1+t2

= 0.319+0.532

= 0.851 sec

FOR TWO

vox = cos 44 degrees * 253m/s = 181.99 m/s

voy = sin 44 degrees * 253= m/s = 175.75 m/s

x = vox * t = 181.99m/s * 5.81s = 1057.36 m

y = voy*t + 1/2gt2

y = 175.75m/s * 5.81s + 1/2 * -9.8m/s2 * (5.81)2 = 855.702 m

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