When a stone is projected with 25
&65° to the horizontal, its heights
are 1 m and 4 m respectively. Then
its range is
(A) 4 m
(B) 8 m
(C) 16 m
(D) 32 m
Answers
Answer:
The answer is a option B
Given : a stone is projected with 25° &65° to the horizontal, its heights are 1 m and 4 m respectively.
To Find : range
Solution:
Velocity = V
projected with 25°
Vertical Velocity = VSin25°
=> Time taken to reach at top = VSin25°/g
V² - U² = 2aS
=> V²Sin²25° = 2g(1)
=> Sin²25° = 2g / V²
Time of Flight = 2VSin25°/g
Horizontal Velocity = VCos25°
=> Range = 2VSin25°/g * VCos25° = 2V²Sin25°Cos25° /g
V²Sin²65° = 2g(4)
=> V²Sin²65° = 8g
=> Cos²65° = 8g/V²
Sin²25° = 2g / V²
=> Sin²25° . Cos²65° = 8g/V² . 2g / V²
=> sin25.Cos25 = 4g/V²
Range = 2VSin25°/g * VCos25° = 2V²(4g/V²) /g = 8 m
range is 8 m
option B is correct
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