A body is thrown vertically upward with speed 40 m/s. Distance travelled in last second of ascent (g=10)
A. 20 m
B. 15 m
C. 5 m
D. 35 m
Answer: Option C
Solution (By JKSSB Mock Tests)
Time of ascent = u/g = 4 s. Distance in last second (4th) = ½g(1)²? Actually, during last second of upward journey, velocity becomes zero. Distance = ½g(1)² = 5 m. Formula: s = u - g/2 (2n-1) with n=4? Let's compute: s_nth = u - (g/2)(2n-1). For n=4 (4th second), s = 40 - 5×7 = 40 - 35 = 5 m. Yes.
Explanation:
Velocity v = fλ constant for given medium. Ultrasound has higher frequency (>20 kHz) than audible sound, so wavelength λ = v/f is smaller. Typical ultrasound in MHz range, λ in mm.
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