Question

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other object will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?

Answer 1

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

$\sqrt{(2H)/g} -> \sqrt{(2 x 179)/ 9.8} = 6.04s$*

Then find xf of projectile $xf= 5.9(6.04) = 37.7$

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

Answer 2

"217 cm" would be the difference throughout the landing locations of the two objects.

Given:

Initial horizontal speed,

• v = 5.9 m/s

Height,

• h = 179 cm

By applying equation of motion, we get

→               $S = ut+\frac{1}{2}at^2$

By substituting the values, we get

→ $179\times 10^{-2}=0+\frac{1}{2}\times 9.8\times t^2$

→            $3.58= 9.8t^2$

                 $t^2 =\frac{3.58}{9.8}$

                  $t = \sqrt{0.36531}$

                     $= 0.61 \ s$

now,

The horizontal distance travelled by the first particle will be:

→ $d = v\times t$

     $= 5.9\times 0.61$

     $= 3.59 \ m$

or,

     $= 359 \ cm$

hence,

The distance between particle will be:

= $d -142$

= $359-142$

= $217 \ cm$

Thus the above approach is right.

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