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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Rest and Motion Kinematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rest and Motion Kinematics

From a point 100 m above the ground, a ball is thrown horizontally...

From a point 100 m100 \text{ m} above the ground, a ball is thrown horizontally with a speed of 20 m/s20 \text{ m/s}. Determine the time it takes to reach the ground in seconds.

Options

  1. A4.474.47
  2. B5.005.00
  3. C4.524.52
  4. D2.002.00

Correct answer

C. 4.524.52

Step-by-step solution

The initial vertical velocity of the ball is uy=0u_y = 0.

The vertical displacement is sy=100 ms_y = 100 \text{ m}.

Using the second equation of motion for the vertical direction:

sy=uyt+12gt2s_y = u_y t + \dfrac{1}{2} g t^2

Substituting the given values and taking g=9.8 m/s2g = 9.8 \text{ m/s}^2:

100=0+12×9.8×t2100 = 0 + \dfrac{1}{2} \times 9.8 \times t^2

100=4.9t2100 = 4.9 t^2

t2=1004.9=100049t^2 = \dfrac{100}{4.9} = \dfrac{1000}{49}

t=10107≈4.5 st = \dfrac{10\sqrt{10}}{7} \approx 4.5 \text{ s}
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