Physic C....help me?

Question

a ball is thrown horizontally from a height of 20m and hits the ground with a speed that is three times its initial speed. What was the initial speed????

Answer 1

Let intial velocity of the ball be u.

In vertically downward direction

uy = 0,s = 20 m, a = g=9.8m/s2

Let time taken to reach the ground be t.

Using second equation of motion,

s = ut +1/2 at2

20 = 0 + 1/2 x9.8 xt^2

t= 2.02 s

Speed on reaching the ground is

v = (u^2+g^2t^2)^1/2 = 3u

u^2+(9.8x2.02)^2=9u^2

u = 7 m/s

Answer 2

y0 = 20

v(y=0) = 3v0

v^2 = v0^2 + 2 a y

you want to solve for v0^2, a = g since acceleration in free fall is due to gravity ... v(0) = v^2 - 2 g y = 3 v0 - 40 g

Answer 3

put the value of gravity you use  as "a"

I've put 10 for gravity, which we use at school

if the initial speed is "u"

when it hits the ground the speed is 3u

in the direction of gravity

S=ut +1/2 at^2

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

t = 2 s

in the direction of gravity

V= u + at

Vy=0 +10 *2

Vy=20 m/s ---------------------------------(1)

horizonally

V=u + at

Vx= u +10*2

Vx=u+20 --------------------------------------(2...

from (1) and (2)

final velocity V

V=sqrt(Vx^2 +Vy^2)                ;V=3u

3u=sqrt [ (u+20)^2 + (20^2) ]

9u^2=u^2 +40u + 800

u^2=5u+100

so

u ={ 5 + sqrt(425)} /2

or

u ={ 5 - sqrt(425)} /2