Answer:
3.467 s
Explanation:
given,
distance , d = 49 mm = 0.049 m
initial speed of the of the rock, v = 17 m/s
time taken by the Heather rock to reach water
using equation of motion
[tex]s = ut +\dfrac{1}{2}at^2[/tex]
taking downward as negative
[tex]-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2[/tex]
4.9 t² + 17 t - 0.049 = 0
now,
[tex]t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}[/tex]
t₁ = -3.47 s , 0.0028 s
rejecting negative values
t₁ = 0.0028 s
now, time taken by the ball of Jerry
using equation of motion
[tex]s = ut +\dfrac{1}{2}at^2[/tex]
taking downward as negative
[tex]-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2[/tex]
4.9 t² - 17 t - 0.049 = 0
now,
[tex]t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}[/tex]
t₂ = 3.47 s ,-0.0028 s
rejecting negative values
t₂ = 3.47 s
now, time elapsed is = t₂ - t₁ = 3.47 - 0.0028 = 3.467 s
