Question

An Alaskan rescue plane traveling 39 m/s drops a package of emergency rations from a height of 198 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s^2. Where does the package strike the ground relative to the point directly below where it was released?

Answer 1

Answer:

The package will strike at 247.65 m.

Explanation:

Given that,

Speed = 39 m/s

Height = 198 m

Acceleration = 9.8 m/s²

We need to calculate the time

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$198=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\dfrac{198\times2}{9.8}$

$t=\sqrt{40.40}$

$t=6.35\ sec$

We need to calculate the distance where the package strike the ground

Using formula of distance

$d= vt$

Put the value into the formula

$d=39\times6.35$

$d=247.65\ m$

Hence, The package will strike at 247.65 m.

Answer 2

y = 0m

y0 = 166m

v0y = 0 m/s

g = 9.8 m/s^2

t = ?

Solve for t:

y = y0 + v0y*t - (0.5)gt^2

0 = 166 - (0.5)(9.8)t^2

t = 5.82 s

Now, using time, we can solve for the range using the equation:

x = vx(t)

x = (40)(5.82)

x = 232.8 m

The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:

v = v0y - gt

v = 0 - (9.8)(5.82)

v = -57.04 m/s