Question

A slingshot fires a pebble from the top of a building at a speed of 15.0 m/s. The building is 30.0 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.

Answer 1

a) horizintally

V_x= 15m/s ( along the x-axis)

time to cover 30 m= = 2.47 seconds apprx

09

V_y= 9.8(2.47) =24.25 m/sw apprx

Resulatant velocity = sqroot( 15^2 + 24.25^2) =28.5 m/s apprx

b) Vertically straight up =

max height reached = v^2/2g = 15^2/19.6=11.5 m apprx

While falling down , the heighjt to be covered= 30 + 11.5= 41.5 m apprx

velocity under free fall = ?2gh= sqroot( 2 x 9.8x 41.5)= 28.5 m/s apprx

c) vertically down:

v^2 - u^2 = 2gh

v^2= 2(9.8)(30) + 15^2

v= 28.5 m/s apprx

Answer 2

a) horizintally

V_x= 15m/s ( along the x-axis)

time to cover 30 m= = 2.47 seconds apprx

V_y= 9.8(2.47) =24.25 m/sw apprx

Resulatant velocity = sqroot( 15^2 + 24.25^2) =28.5 m/s apprx

b) Vertically straight up =

max height reached = v^2/2g = 15^2/19.6=11.5 m apprx

While falling down , the heighjt to be covered= 30 + 11.5= 41.5 m apprx

velocity under free fall = √2gh= sqroot( 2 x 9.8x 41.5)= 28.5 m/s apprx

c) vertically down:

v^2 - u^2 = 2gh

v^2= 2(9.8)(30) + 15^2

v= 28.5 m/s apprx

Answer 3

a)

Be aware, the X direction speed never changes but the Y direction does.

vf^2 = vi^2 + 2ad

vf^2 = 0^2 + 2(9.8 x 30)

vf = 24.248 m/s



b) First we'll calculate how far the ball goes up before it comes to rest.

02 = 225 + 2(-9.8 x d)

Solve for d which is 11.479 m

Now do 30 +11.479 to get new height or d

vf2 = 02 + 2(9.8 x 41.479)

vf = 28.51 m/s

c)

vf2 = 225 + 2(9.8 x 30)

vf = 28.51 m/s