Interactive solution 6.39 presents a model for solving this problem. a slingshot fires a pebble from the top of a building at a speed of 13.0 m/s. the building is 23.0 m tall. ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

Given that

u = 13 m/s

h= 23 m

a) horizontally

We know that acceleration due top gravity is in vertical direction.

v²=u'²+ 2 g h    ( u'= 0 )

v² = 2 x 9.81 x 23

v= 21.24 m/s  ( vertical direction)

In the horizontal direction

u = 13 m/s

The resultant velocity

V²=v²+u²

V²=21.24²+13²

V= 24.91 m/s

(b) vertically straight up

u= 13 m/s

v²=u²+ 2 g h

v²=13²+ 2 x 9.81 x 23

v= 24.91 m./s

c)vertically straight down.

u= 13 m/s

v²=u²+ 2 g h

v²=13²+ 2 x 9.81 x 23

v= 24.91 m./s

we need to know the time interval during which the force acts. impulse of force = momentum change