a body projected at angle 30 to the horizontal so as just to clear two walls of height 5 m and 10 m apart from each other find total range of projectile
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Correct option is
B
70 m
Horizontal component of projectile's motion
2.5
50
=20㎧
To have a range of 50m, maximum height would be reached in
2
2.5
=1.25secs (neglecting walls for now)
1.25×g=initial velocity=12.5㎧
Height attained over walls=
2g
V
2
=
2g
(12.5)
2
=7.8125m
Adding wall weight(7.812+7.5)=15.3125m
above ground level
So initial vertical v component from ground level
=
2gh
=
2×10×15.3125
=17.5㎧
Time to maximum height from the ground=
g
V
=
10
17.5
=1.75sec
Total time in air=(1.75×2)=3.5sec
Range=3.5×20
=70m
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