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NEET PHYSICSMedium

A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s10 \text{ m/s}, then maximum height attained by it is (Take g=10 m/s2g = 10 \text{ m/s}^2):

A

8 m

B

10 m

C

12 m

D

16 m

Step-by-Step Solution

  1. Define Maximum Height (HH): At maximum height, the final velocity v=0v = 0. Let the initial velocity be uu. Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as with a=−ga = -g: 02=u2−2gH  ⟹  u2=2gH…(1)0^2 = u^2 - 2gH \implies u^2 = 2gH \quad \dots(1)
  2. Analyze Motion at Half Height (H/2H/2): The problem states that at height s=H/2s = H/2, the velocity is v=10 m/sv = 10 \text{ m/s}. Apply the kinematic equation for this point: v2=u2−2g(H2)v^2 = u^2 - 2g\left(\frac{H}{2}\right) (10)2=u2−gH(10)^2 = u^2 - gH
  3. Substitute and Solve: Substitute u2=2gHu^2 = 2gH from equation (1) into the second equation: 100=2gH−gH100 = 2gH - gH 100=gH100 = gH
  4. Calculate HH: Given g=10 m/s2g = 10 \text{ m/s}^2: 100=10×H100 = 10 \times H H=10 mH = 10 \text{ m} .
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