We read your photo as:

सामाजिक अध्ययन (समय : ४० मिनेट) प्रश्नावली (क) १. जन्म र मृत्युदर घटना कम गर्न उपायहरु बताउनुहोस्। (ख) जलवायु परिवर्तन तpai. तलका घटनाको कारण बताउनुहोस्: (१) बाढि (२) पहिरो (३) मानब निर्मित प्रदूषण (ग) भूकम्पबाट हुने क्षति कम गर्न अपनाईएका उपायहरु लेख्नुहोस्।

Physics

ScanSolve AI Answer

Step-by-step solution

8 steps
  1. For maximum height, final velocity v=0v=0 m/s.

    v2=u2−2ghv^2 = u^2 - 2gh

    Substitute u=20u=20 m/s, g=10g=10 m/s2^2:

    0=(20)2−2×10×h0 = (20)^2 - 2 \times 10 \times h 400=20h400 = 20h h=20 mh = 20\ \mathrm{m}

    20\ m

    प्रश्न (2)

  2. Take upward as positive direction from top of tower.

    Displacement to ground s=−50s=-50 m, u=20u=20 m/s, a=−g=−10a=-g=-10 m/s2^2.

    s=ut+12at2s = ut + \frac{1}{2}at^2

    Substitute values:

    −50=20t+12(−10)t2-50 = 20t + \frac{1}{2}(-10)t^2 −50=20t−5t2-50 = 20t - 5t^2
  3. Rearrange into standard quadratic equation.

    5t2−20t−50=05t^2 - 20t - 50 = 0

    Divide by 5:

    t2−4t−10=0t^2 - 4t - 10 = 0
  4. Solve using quadratic formula t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=1a=1, b=−4b=-4, c=−10c=-10.

    t=4±(−4)2−4(1)(−10)2(1)t = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-10)}}{2(1)} t=4±16+402=4±562=4±2142=2±14 st = \frac{4 \pm \sqrt{16 + 40}}{2} = \frac{4 \pm \sqrt{56}}{2} = \frac{4 \pm 2\sqrt{14}}{2} = 2 \pm \sqrt{14}\ \mathrm{s}
  5. Take positive root (time cannot be negative)

    t=2+14t = 2 + \sqrt{14} s.

    2 + \sqrt{14}\ s

    प्रश्न (3)

  6. Time to reach maximum height t=ugt = \frac{u}{g}.

    Substitute u=19.6u=19.6 m/s, g=9.8g=9.8 m/s2^2:

    t=19.69.8=2 st = \frac{19.6}{9.8} = 2\ \mathrm{s}

    2\ s

    प्रश्न (4)

  7. Time to fall from height 50 m with initial vertical velocity 0 m/s.

    h=12gt2h = \frac{1}{2}gt^2 50=12×10×t250 = \frac{1}{2} \times 10 \times t^2 50=5t250 = 5t^2 t2=10t^2 = 10 t=10 st = \sqrt{10}\ \mathrm{s}
  8. Horizontal distance (range) R=uxtR = u_x t, where horizontal velocity ux=10u_x=10 m/s (constant).

    R=10×10=1010 mR = 10 \times \sqrt{10} = 10\sqrt{10}\ \mathrm{m}

    10\sqrt{10}\ m

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question