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

v₀= 19.23 m/s

Step-by-step 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