Question

A stone is dropped into a river from a bridge 44.0 m above the water. Another stone is thrown vertically down 1.72 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone?

Answer 1

Answer:

u₀ = 17.14 m/s

Explanation:

given,

bridge height = 44 m

initial speed of the first stone = 0 m/s

initial speed of the second stone = ?

difference after which the second stone is thrown = 1.72 s

for stone 1

$h = ut + \dfrac{1}{2}gt^2$

$h =\dfrac{1}{2}gt_1^2$

for stone 2

$h = u_0 (t_1-t) + \dfrac{1}{2}g (t_1-t) ^2$

$t_1 =\sqrt{\dfrac{1}{2}gh}$

$t_1 = \sqrt{\dfrac{1}{2}\times 9.81\times 44}$

t₁ = 14.69 s

$44 = u_0 \times 1.72 + \dfrac{1}{2}g\times 1.72 ^2$

u₀ = 17.14 m/s