Physics Projectile question?

Question

a helicopter is flying horizontally @ 20m/s and at a height of 180 . a small object falls from it. calculate the time for it to reach the ground and the horizontal distance that it travels before hitting the ground?

plz seroiud answers only and plz do explain ur method so that i won't have to trouble anyone again on these sorts of questions.

it's 180m and 20m/s and 9.8m/s^2

so we use d=st for horizontal distance that the object travelled before hitting the ground? then how come we don't use s=ut+0.5xaxt^2 equation to find the horizontal distance?
where s is displacement, u is intial velocity, a is accleration and t is time.

Answer 1

In the vertical direction earth is pulling the object downward with a force.

And all objects are pulled down by the earth with an acceleration given by g = 9.8 m/ s^2.

First we consider the vertical motion alone.

In the vertical direction
the initial velocity u = 0.
the acceleration a = g.
The time t = t to be found.
Displacement in the vertical direction s = 180 m.

We use the equation s = ut + 05 a tt.

Using this we find t = 6 second.

Now let us consider the horizontal motion.

The object is released from the plane which was moving with a speed of 20m/s.

There fore all objects which was in the plane will also have a speed of 20m/s in the horizontal direction.

Hence in the horizontal direction the initial speed is 20m/s.

The acceleration in this direction is zero since no force is acting in this direction.

The time t is the same as 6 second.

Distance we have to find.

We use the equation s = ut + 05 a tt.

Since a = 0 the equation reduces to s = ut. = 20 x 6 = 120m

The horizontal displacement is 120 m.

Answer 2

You can use the following formula
s=ut +0.5ft*t

where s= distance, u is initial velocity f=accleration t= time

Consider only verical motion. In this case u=0 and f= g (acceleration due to gravity g)
s= 0.5gt*t
or t= square root of 2s/g

s=120 somethings (metres or feet ?)

t=240/g (where g is in the appropriate units)

Now consider horizontal motion the object moves horizontally at 20 m/s for time =240/g

V=d/t or d=vt therefore d (distance travelled horizontally) =20.240/g metres

It's actually less than this because of viscosity but your teacher won't be concerned about this effect.

Answer 3

t: time, e: distance, a: acceleration, g: acceleration due to gravity,
v: velocity.

time of fall:
t = sqr(2e/a) = sqr(2e/g) = sqr(2 * 180 / 9.81) = 6.058s

distance travelled horizontally:
e = vt = 20 * 6.058 = 121.156m
( a bit less, due to air resistance that will slow down the horizontal velocity, but that is not your concern at this time)

Velocity when hitting the ground:
v = at = 9.81 * 6.058 m/s = 60m/s (but this is not asked for).
(here again, ignoring air resistance).

Answer 4

initially vertical velocity of body is zero
so h=0.5*g*t^2
so t=6 seconds
horizontal coponent of velocity remains unaltered( acc is only vertical)
so horizontal dist = v* t
t= 6 sec (from above)
so dist = 20*6 = 120 metres