Question

Heather and Jerry are standing on a bridge 49 mm above a river. Heather throws a rock straight down with a speed of 17 m/sm/s . Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance.
A) How much time elapses between the first splash and the second splash?

Answer 1

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

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = -17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² + 17 t - 0.049 = 0

now,

$t_1 = \dfrac{-(17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

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

$s = ut +\dfrac{1}{2}at^2$

taking downward as negative

$-0.049 = 17 t -\dfrac{1}{2}\times 9.8\times t^2$

4.9 t² - 17 t - 0.049 = 0

now,

$t_2 = \dfrac{-(-17)\pm \sqrt{17^2 - 4\times 4.9 \times (-0.049)}}{2\times 4.9}$

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