physics projectile motion?

Question

A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45°. With what speed was it thrown?
14 m/s
20 m/s
28 m/s
32 m/s
40 m/s

can you show the working out thx

Answer 1

it stikes ground at angle of 45 degrees means:
v(y)/v(x)=tan45=1
v(y)=v(x)
v(x) remains constant throughout
v(y)=(2gh)^1/2
g=10 , h=20m
hence v(y)=v(x)=(2gh)^1/2 put all the values and
you willget v(y)=20m/s

Answer 2

20 m/s
since the ball hit the ground at 45 degrees, its vertical and horizontal components are equal when hitting the ground. the hor. speed is constant, and equal to the vertical speed when the ball hits the ground.
First find how long it would take for the ball to hit the ground from the formula
height=0.5*g*t^2, where g is about 10
20=0.5*10*t^2
t^2=4
t=2
Now you can find the vertical speed on the ground;
V=g*t
V=10*2=20m/s
The vertical speed equals that too on the ground and si nce it does not change throughout the motion, it was 20m/s at the beginning too.
Good luck with physics, I always hated it

Answer 3

28 m/s, just use an angular