Mapt A mortar crew is positioned near the top of a steep hill. Enemy forces are
ID: 1880858 • Letter: M
Question
Mapt A mortar crew is positioned near the top of a steep hill. Enemy forces are charging up the hill and it is necessary for the crew to spring into action. Angling the mortar at an angle of 50 0° (as shown), the crew fires the shell at a muzzle velocity of 203 feet per second. How far down the hill does the shell strike if the hill subtends an angle 36.00 from the horizontal? (Ignore air friction.) Number 1n How long will the mortar shell remain in the air? Number How fast will the shell be traveling when it hits the ground? Number mis The mortar is like a small cannon hat launches shelis at steep angles O Prevos G.ve, up & View Sohtion ( Check Answer Ext NetExplanation / Answer
Horizontal range
d cos 36= 203 cos 50* t
d= (203 cos 50/cos 36)*t
d= 161.29 t ...(i)
Considering motion along vertical
- d sin 36= 203 sin 50*t - 0.5*32.2*t^2
Putting value of d
- 161.29*sin 36*t= 155.5t - 16.1 t^2
t= 15.547 seconds
Putting in ..(i)
d= 2507.58 ft
==========
a)
Vertical distance covered
h= d sin 36= 1474 ft
==========
b)
Time of flight,
t= 15.547 second
========
c)
Vertical velocity at this instant of time
Vy= 203 sin50 -32.2*15.547 = - 345.1 ft/s
Horizontal velocity
Vx= 203 cos 50= 130.46 ft/s
Net velocity
V^2= Vx^2+ Vy^2
= 368.94 fts
======
Comment in case any doubt.. good luck