the ceiling of a long hall …
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Gaurav Seth 5 years, 2 months ago
Given,
Hight of Hall( H) = 25 m
Let ball is thrown with speed 40m/s at an angle ∅ with horizontal .
We know,
H = u²sin²∅/2g
25= (40)² × sin²∅/2× 10
25 = 80 × sin²∅
Sin²∅ = 25/80
Sin∅ = 5/4√5
Cos∅ = √55/4√5
Now, horizontal range = u²sin2∅/g
= (40)²× 2sin∅×cos∅/g
= 160 × 2 × 5/4√5 × √55/4√5
= 20√55 m
1Thank You