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A ball is dropped from a height. What is the height of the building if it takes 4 seconds to reach the ground? Also, what is the initial speed required to throw a ball to a maximum height of 100m?

Physics

ScanSolve AI Answer

Answer

D

Step-by-step solution

9 steps

  1. Identify the given values and the formula for free fall.

    For an object dropped from rest, the initial velocity u=0u = 0 m/s.
    The time taken to reach the ground t=4t = 4 s.
    The acceleration due to gravity g=9.8g = 9.8 m/s2^2.
    The formula for the distance (height) is h=ut+12gt2h = ut + \frac{1}{2}gt^2.

  2. Substitute the values into the formula and calculate the height.

    h=(0 m/s)(4 s)+12(9.8 m/s2)(4 s)2h = (0 \text{ m/s})(4 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s}^2)(4 \text{ s})^2 h=0+12(9.8 m/s2)(16 s2)h = 0 + \frac{1}{2}(9.8 \text{ m/s}^2)(16 \text{ s}^2) h=(4.9 m/s2)(16 s2)h = (4.9 \text{ m/s}^2)(16 \text{ s}^2) h=78.4 mh = 78.4 \text{ m}

    The height of the building is 78.4 m.

    The correct option is (d).

    d) 78.4 m\boxed{\text{d) } 78.4 \text{ m}}

  3. Identify the given values and the formula for vertical motion.

    The maximum height attained h=100h = 100 m.
    At the maximum height, the final velocity v=0v = 0 m/s.
    The acceleration due to gravity g=−9.8g = -9.8 m/s2^2 (negative because it opposes the upward motion).
    The formula relating initial velocity, final velocity, acceleration, and displacement is v2=u2+2ghv^2 = u^2 + 2gh.

  4. Substitute the values into the formula and solve for the initial speed uu.

    02=u2+2(−9.8 m/s2)(100 m)0^2 = u^2 + 2(-9.8 \text{ m/s}^2)(100 \text{ m}) 0=u2−1960 m2/s20 = u^2 - 1960 \text{ m}^2/\text{s}^2 u2=1960 m2/s2u^2 = 1960 \text{ m}^2/\text{s}^2 u=1960 m2/s2u = \sqrt{1960 \text{ m}^2/\text{s}^2} u≈44.27 m/su \approx 44.27 \text{ m/s}

    Rounding to one decimal place, the initial speed is 44.3 m/s.
    Comparing with the options, 44.2 m/s is the closest.

    The correct option is (b).

    b) 44.2 m/s\boxed{\text{b) } 44.2 \text{ m/s}}

  5. Identify the given values and the formula for weight.

    The mass of the man m=60m = 60 kg.
    The acceleration due to gravity on Earth gE=9.8g_E = 9.8 m/s2^2.
    The acceleration due to gravity on the Moon gM=gE6g_M = \frac{g_E}{6}.
    The formula for weight is W=mgW = mg.

  6. Calculate the acceleration due to gravity on the Moon.

    gM=9.8 m/s26≈1.633 m/s2g_M = \frac{9.8 \text{ m/s}^2}{6} \approx 1.633 \text{ m/s}^2

  7. Calculate the weight of the man on the Moon.

    WM=m×gMW_M = m \times g_M WM=60 kg×1.633 m/s2W_M = 60 \text{ kg} \times 1.633 \text{ m/s}^2 WM=97.98 NW_M = 97.98 \text{ N}

    Rounding to the nearest whole number, the weight is approximately 98 N.

    The correct option is (d).

    d) 98 N\boxed{\text{d) } 98 \text{ N}}

  8. Identify the given values and the formula for weight.

    The weight of the object on Earth WE=196W_E = 196 N.
    The acceleration due to gravity on Earth gE=9.8g_E = 9.8 m/s2^2.
    The formula for weight is W=mgW = mg.

  9. Rearrange the formula to solve for mass and substitute the values.

    m=WEgEm = \frac{W_E}{g_E} m=196 N9.8 m/s2m = \frac{196 \text{ N}}{9.8 \text{ m/s}^2} m=20 kgm = 20 \text{ kg}

    The mass of the object is 20 kg.

    20 kg\boxed{20 \text{ kg}}
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