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

u₀ = 17.14 m/s

Step-by-step 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 + (1)/(2)gt^2`

`h =(1)/(2)gt_1^2`

for stone 2

`h = u_0 (t_1-t) + (1)/(2)g (t_1-t) ^2`

`t_1 =\sqrt{(1)/(2)gh}`

`t_1 = \sqrt{(1)/(2)* 9.81* 44}`

t₁ = 14.69 s

`44 = u_0 * 1.72 + (1)/(2)g* 1.72 ^2`

u₀ = 17.14 m/s