Question

Two students are messing around with a nerf gun during a fire drill. They manage to fire a foam dart directly upwards from an inital height of 1.5 m above the ground. Four seconds later, the foam dart hits the ground. What was the initial velocity of the dart?

Answer 1

Answer:

v₀= 19.23 m/s

Explanation:

Look at the attached graphic

Foam dart Kinematic 1-2 (upward movement), vf₂=0 ,

We calculate t₁ and y₁ to reach the highest point (2)

vf₂=v₀-gt₁  

t₁=v₀/g

v f₂²=v₀²-2g*y₁

2g*y₁ =v₀² , y₁=v₀²/2g

Foam dart Kinematic 2-3 (downward movement), v₀₂=0 ,

We calculate t₂ and y₂ from the highest point (2) to

touch the ground (3)

vf₃=v₀₂+gt₂   , v₀₂=0

gt₂=vf₃ , t₂=vf₃/g

vf₃²=v₀₂²+2g*y₂

$v_{f3} =\sqrt{2*g*y_{2} }$

y₂= 1.5+y₁

y₂= 1.5+v₀²/2g

$v_{f3} =\sqrt{2g(1.5+v_{o}^{2}/2g) }$

$vf_{3} =\sqrt{v_{o}^{2}+29.4 }$

We propose the equation for the total time (Four seconds) :

t₁ = time it takes from position 1 to position 2 (going up)

t₂: and time that takes from position 2 to position 3 ( going down)

t₁+t₂=4

v₀/g+ vf₃/g =4

v₀ + vf₃ =4g

$v_{o} +\sqrt{v_{o} ^{2}+29.4 } =4*g$  : we move v to the other side and square both sides of the equation

v₀²+29.4=(4g-v₀²)²

v₀²+29.4=(4g)²-8gv₀+v₀² we eliminate v₀²

29.4=(4g)²-8gv₀

8*g*v₀=(4*g)²-29.4

v₀=1507.24/78,4

v₀= 19.23 m/s