Physics Question Please Answer!?

Question

A package is droppes from an airplane 344 m above the ground. If air resistance is neglected how long will it take the package to reach the ground?

Please I need it for Physics/.

Answer 1

You know that the acceleration of any body falling on earth is always 9.8 m/s irrespective of its mass.

We have height or distance travelled = 344

Now the motion of package is uniformly accelerated motion.

At initial time when it was dropped, its velocity was 0.

Let s, u, g and t be distance, initial velocity, acceleration due to gravity and time taken to reach earth respectively.

By using second equation of motion which is

s = ut + 1/2 at^2, we have

344 = 0 + 1/2 (9.8)t^2

from where you can solve for t

So

t = sqrt of {2(344)/9.8} = sqrt {70.204081632653061224489795918367}

= 8.3787875992086743685842805982713 seconds.

which is approximately 8.3 seconds.

am I right???

Answer 2

using v^2=u^2+2ax

v^2=0+2x9.8x344= 6742.4

after square rooting 6742.4 you get 82.117

again using x=1/2(u+v)t  we get after some substitution

344=t+(0+82.117)

thus to find t we say

t=x/v  which is

t=344/82.117=4.1891 seconds

this is assuming the acceleration due to velocity is 9.8

Answer 3

use the equation s = ut + 1/2 at^2

s = 344

u = 0

a =10

solve for t

answer= 8.3s