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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 )

Arun
25763 Points
3 years ago
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)
= $\sqrt$6400+3600 = $\sqrt$10000
= 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.