Question

A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance of 155 m above the surface of a moon. The rate of freefall acceleration on this airless moon is 2.79 m/s2. (a) How long does it take for the box to reach the moon's surface? (b) What is its horizontal displacement during this time? (c) What is its vertical velocity when it strikes the surface? (d) At what speed does the box strike the moon?

Answer 1

Answer:

Answered

Explanation:

horizontal velocity v= 27.0 m/s

height of space craft = 155 m

rate of free fall acceleration on this airless moon is 2.79 m/s^2.

we know that

a) S= V×t+ 0.5at^2

h= 0×t+ 0.5×2.79t^2               [initial vertical velocity is zero]

155= 0.5×2.79t^2

⇒ t= 10.54 secs

this the the time that the box will take to reach the moon.

b) the horizontal velocity v= 27 m/s and t= 10.54 sec

therefore horizontal displacement d= 10.54×27= 284.58 m

c) for final vertical velocity

we can use the formula

(v_final)^2=( v_initial)^2+ 2×a×h

(v_final)^2=0^2+ 2×2.79×155

v_final= 29.40 m/s [this its  vertical velocity when it strikes the surface]

d) speed with which the box strike the moon will be resultant of the final vertical and horizontal velocity vectors that

$\sqrt{27^2+29.40^2}$

= 39.91 m/s

Answer 2

Answer:

a) 10.54 sec

b) 284.58 m

c) 29.406 m/s

d) 39.92 m/s

Explanation:

Given data:

velocity of spacecraft = 27.0 m/s

rate of free fall acceleration is 2.79 m/s^2

distance of moving aircraft from mooon surface is 155 m

a. from kinematic eqaution of motion we have

$y = Vi\times t + (\frac{1}{2}) a\times t^2$

where y = 155 m

           Vi = 0  as this relation  is for vertical motion, so the 27.0 m/s is not included

and a = 2.79 m/s^2.

Solving for t we get

t = 10.54 sec

b.

we know that $V = \frac{d}{t}$

$d = v\times t$

   $= 27 \times 10.54 = 284.58 m$

c.  from the kinematic formula

v = u + at

$v = 0 + 2.79\times 10.57$

v = 29.4066 m/a

d. $v = \sqrt { 27^2 + 29.406^2}$

     v = 39.92 m/s