I got it wrong /2018 12:45 PM 39.1/1009/19/2018 11:16 AM Print on 11 of 11 Incor

Question

I got it wrong

/2018 12:45 PM 39.1/1009/19/2018 11:16 AM Print on 11 of 11 Incorrect Sapling Learning A mortar" crew is positioned near the top of a steep hil. Enemy Soroes are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of 8 670 (as shown), the crew fires the shell at a muzzle velocity of 144 feet per second. How far down the hil does the shell strike if the hill subtends an angle 350P from the on.) 156 m. How long will the mortar sheil remain in the air? 9.10 How fast will the shel be traveing when hits the ground? 51.7

Explanation / Answer

Horizontal range

d cos 35= 144 cos 67* t

d= (144 cos 67/cos 35)*t

d= 68.69 t ...(i)

Considering motion along vertical

- d sin 35= 144 sin 67*t - 0.5*32.2*t^2

Putting value of d

- 68.69*sin 35*t= 132.55 t - 16.1 t^2

t= 10.68 seconds

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a)

Putting in ..(i)

d= 733.61 ft = 223.6 m

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b)

Time of flight,

t= 10.68 second

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c)

Vertical velocity at this instant of time

Vy= 144 sin67 -32.2*10.68 = - 211.343 ft/s

Horizontal velocity

Vx= 144 cos 67= 56.265 ft/s

Net velocity

V^2= Vx^2+ Vy^2

= 218.7 ft/s

= 66.66 m/s.

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Comment in case any doubt.. good luck

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