A body is projected into air with some velocity so that it just clears two walls of height 10m and 5m at distances 5m and 10m respectively from the point of projection. Ig = 10 m/s 1. The velocity or projection is es a) 5 b) 18 m/s c) 16 m/s d) 18 m/s
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Answered by
0
Answer:
70 m
Explanation:
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
=70mv
Answered by
0
Answer:
The velocity or projection is b) 18 m/s
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