Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to
2 answers:
Answer:
B. elastic PE to kinetic energy
Explanation:
As we know that when we stretch the slingshot to some distance then the elastic energy is stored into the sling which is given as
here we know that
k = elastic constant of the sling
x = stretch in the sling
now when we release the sling from that position then the shot will move with certain kinetic energy from it and sling comes to its natural length
so here we can say that elastic potential energy of the sling will convert into kinetic energy of the shot here so correct answer will be
B. elastic PE to kinetic energy
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct