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A shell is fired from a gun and has a horizontal range of 960m and time of flight of 12s. Find the magnitude and direction of the velocity of projection. (take )

patience , 8 Years ago
Grade 11
anser 1 Answers
Arun
using distance = velocity * time
for horizontal motion
960 = U* 12
U = 80 m/sec
 
using  s = ut + ½ a t2
 for vertical motion
0 = V* 12 – ½ *10* 122
V = 60 m/sec
 
speed of projection = \sqrt(u2 + v2)
 = \sqrt6400+3600 = \sqrt10000
 = 100 m/sec.
 
the angle of projection is \theta above the horizontal
tan\theta = 60/80 = ¾
\theta = 36.87
 
hencethe shell is projected at 100 m/secat an angle of 36.87 degree above the horizontal.
Last Activity: 8 Years ago
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