Question

A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 24.0 m/s, and the air drag on it is 0.266 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground

Answer 1

Answer

given,

weight of the stone , W = 5.29 N

initial speed of the stone, v = 24 m/s

Air drag = 0.266 N

net force acting on the stone when it reaches maximum height is zero

$F = F_{drag} + W$

$F = 0.266 + 5.29$

     ma = 5.56 N

$a =\dfrac{5.56}{\dfrac{5.29}{9.8}}$

  a = 10.3 m/s²

maximum height of the stone

    $H = \dfrac{u^2}{2g}$

    $H = \dfrac{24^2}{2\times 9.8}$

         H = 29.38 m

b) from the maximum height stone will fall as free fall body

using equation of motion

  v² = u ² + 2 a h

   u = 0 m/s

  $v^2= 0^2 + 2 \times\dfrac{w-F_{air}}{m} \times 29.38$

  $v^2=2\times\dfrac{5.29-0.266}{\dfrac{5.29}{9.8}} \times 29.38$

  v = √546.89

  v = 23.39 m/s

speed at which ball hit the ground is equal to 23.39 m/s