Question

While sitting on a tree branch 11 m above the ground, you drop a chestnut. When the chestnut has fallen 2.6 m , you throw a second chestnut straight down.

What initial speed must you give the second chestnut if they are both to reach the ground at the same time?

Answer 1

here,

h1 = 11 m

h2 = 2.6 m

for the dropped stone

let the time taken be t1 for h1

using second equation of motion

h1 = 0 + 0.5 * g * t1^2

11 = 0 + 0.5 * 9.81 * t1^2

solving for t1

t1 = 1.498 s

let the time taken be t2 for h2

using second equation of motion

h2 = 0 + 0.5 * g * t2^2

2.6 = 0 + 0.5 * 9.81 * t2^2

solving for t2

t2 = 0.728 s

for the stone throwed with a initial speed 'u'

the time taken by throwred stoen , t = t1 - t2 = 0.77 s

for vertical direction

h1 = u * t + 0.5 * g * t^2

11 = u * 0.77 + 0.5 * 9.81 * 0.77^2

solving for u

u = 10.51 m/s

the initial speed is 10.51 m/s