please help! A slingshot fires a rock from the top of a building at a speed of 13.0m/s...?

Question

A slingshot fires a rock from the top of a building at a speed of 13.0m/s. The building is 20m tall, ignoring air resistance, find the speed with which the rock strickes the ground when the rock is fired horizontally?

Could someone please expalin how to answer this, Thank you! :-)

Answer 1

take initial speed(u)=13i + 0j

acceleration= -gj or -10m/s^2

displacement(s)= xi - 20j

s=ut + 1/2 at^2(take in y direction)

-20= 0t +1/2 *-10t^2

20*2/10=t^2

t^2=4

t=2 second

v= u + at (v is final velocity)

v= 0 -10 *2 [in y direction]

v=-20m/s

the total v= 13i - 20j

v=23.8 m/s[approximately]

Answer 2

On the assumption the rock is fired horizontally.Find the time the rock takes to reach the ground from the equation distance of drop

(d)=I/2(gt^2) take g = 10(in usual units) we get 20 = (10 *T^2)2 so

40 = 10 t^2 ; then T^2 =4 so time of drop = root 4=2 seconds.

Final vertical velocity is gt = 10 x 2 = 20 m/s. At the end of flight we horizontal component of speed =13m/s and vertical velocity = 20m/s.So from a triangle of velocities we get rock velocity = root(20squared = 13 squared) =23.85m/s(approx because g is not=10)

Answer 3

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Answer 4

It doesnt chande the speed of the rock vertically if u shoot it horizontally or just drop it.