Energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ignoring all frictional effects, find the speed with which the pebble strikes the ground

Horizontal velocity is 14 m/s

Vertical velocity is 28.3 m/s

Explanation:

Hello dear friend, you have not mentioned the type of speed that you require for this problem.

In this case, there are three possibilities with which the slingshot can be fired i.e. Horizontally, Vertically straight up and vertically straight down. Below is the explanation / answer to all three possibilities

Fired horizontally:

initial conditions:  

Vertical Velocity = 0 ; Horizontal Velocity = 14m/s

final conditions:

Vertical Velocity  (v²  = u² + 2gs) but initial vertical velocity is zero

v²  =  2gs so v²  = 2(9.8)(31) = 607

v = 24.6m/s

but Horizontal Velocity is still  = 14m/s

Resultant velocity from these two velocity components (Pythagoras theorem)

V² = (v horizontal)² + (v vertical)² = 14² + 24.6²

V = 28.3m/s

angle = tan ⁻¹(24.3/14) = 60.1⁰

V = 28.3m/s at angle of 60.1⁰ to the horizontal

Fired Vertically  Straight Up

distance before the pebble reaches maximum height from top of building

v²  = u² + 2gs

where, v is zero at maximum height

g is minus for upward motion.

v²  = u² + 2gs

0  = 14² -  2(9.8)s

s = 196/19.6 = 10.0m

totals distance from maximum height to the ground = 10.0 m + 31.0 m = 41.0m

v²  = u² + 2gs

now u from maximum height is 0 and g is positive for downward motion

v²  =  2gs

v²  =  2(9.8)(41.0)

v = 28.3m/s

v = 28.3m/s vertically straight up

Fired Vertically Straight Down

v²  = u² + 2gs

u = 14m/s, g = 9.8m/s², s = 31.0m

v²  = 14² + 2(9.8)(31.0)

v = 28.3m/s

v = 28.3m/s vertically straight down

an energy transformation flow diagram is shown

where is it ?

conduction, convection, radiation