another projectile motion question?

Question

A steel ball rolls down a ramp (like a "ski jump"), leaves the ramp in the horizontal direction with 0.829m/s and falls to the ground which is 91.2cm below the ramp. How far in the horizontal direction does the ball move ?

the answer is 0.357464, but can someone please do it step by step

Answer 1

acceleration in the horixontal direction is 0. nothing is acting on the ball.. so the velocity throughout the action is constant in the horizontal direction.

if it leaves the ramp with a velocity ONLY in the horizointal direction, that means that it has an initial velocity of 0 in the vertical.

so Vo = 0m/s in the vertical direction
since gravity acts on it in the vertical direction, you know the initial velocity and the distance travelled, you can use the formula: x = Vit + 1/2(at^2)
to find the time it takes.
or:
91.2cm = 0.912m
0.912 = 0(t) + (1/2)(9.81)t^2
(0.912)(2)/9.81 = t^2
t = 0.4311s

since the ball is travelling at 0.829m/s horizontally for 0.4311s.
the ditance travelled is:
0.829m/s * 0.4311s = 0.357464 m

Answer 2

The horizontal velocity doesn't change so the horizontal distances is = 0.829 (m/s) x time (seconds) = m

The inital vertical velocity is zero "leaves the ramp in the horizontal direction".

So the ball falls 91.2 cm or 0.912 m

It is acelerated by gravity at 9.8 m/s^2 = g

The distance formula is

distance = 1/2 x g x time^2

0.912 = (1/2)(9.8) (time)^2

time = sqrt ( 0.912*2/9.8)

Plug time into the first equation can calculate distance.

Answer 3

From h = 1/2*g*t^2 t = sqrt(2*h/g) this the time fall,
then x= v*t, v= 0,828 m/seg. horizontal distance.
g = acceleration of gravity 9,80 m/seg^2
h = 0,912 m.
you do the calculations.

Answer 4

i am not taking the exact values but can show you the method
time of fall =sqrt(2h/g)
hz. distance=horizontal speed *time
so time comes out to be round 41sec
and distance .357m