an airplane is travelling in 180kmh, it drops a supply to a land 160m below.?

Question

the supply should be dropped how many seconds b4 the plane is directly overhead?

Answer 1

We first have to take the time it would reach the ground. We can use

yf = yi +viy*t +0.5 a*t*t

yi=height of plane = 160 m
yf = ground level = 0 m
viy = velocity of package on the vertical component= 0 m/s(it was dropped)
a = gravitational acc = -9.8 m/s/s

Plug and play:
0 = 160 + (0) t + 0.5 (-9.8) *t*t

or

0 = 160 - 4.9 t *t
160 = 4.9 t*t
t*t = 160/4.9

t squared = 32.653061224489795918367346938776

finally:
t = 5.7142857142857142857142857142857 s

Now, this is the time it takes to drop. This is also the amount of time it takes to go forward from the point it was dropped to the point the plane is overhead.

So the answer is 5.71 seconds.

Answer 2

Since the speed of the plane 180 km/h horizontally while the vertical speed is zero. The supply vertical speed will be zero.

S=Distance traveled= 160 meter
U= Initial velocity
g= Acceleration=9.8m/sec^2
t= time
Using the formula
S =Ut +1/2 gt^2
160 = (0)t +1/2 (9.8)(t^2)
t^2=320/9.8
t= sq rt of(320/9.8)
t=5.714275714=5.7 seconds

Answer 3

First, use y = 0.5*gt^2 with your known y and constant g to find t. Then, multiply by the given x-velocity of 180 kmh to find the required lead distance. This ignores air resistance in both dimensions.

Answer 4

AS THE PLANE HAS NO VERTICAL VELOCITY,


160= 0.t+1/2. gt^2
= 4.9 t^2

t= 5.71 seconds

Answer 5

the speed=180kmph=180*5/18=50m/s
time taken=horizontaldits/horizontal speed=160/50=3.2 sec
it should be dropped 3.2 sec before the plane isdirectly overhead

Answer 6

exactly the time it takes the supply to reach the ground, use the formula 9.81/2*t^2=160

Answer 7

DISTANCE = 1/2 G T(SQUARED)

160 = 1/2 X 9.81 T (SQUARED)

T = 5.71 SECONDS