Projectile motion physics question 11 please
1 answer:
Answer:
tair = 11.5 s
vi = 60.3 m/s
Explanation:
In the x direction:
Given:
x₀ = 0 m
x = 600 m
v₀ = v cos 30° = v √3/2
a = 0 m/s²
x = x₀ + v₀ t + ½ at²
600 = 0 + (v √3/2) t + ½ (0) t²
600 = √3/2 v t
1200/√3 = v t
In the y direction:
Given:
y₀ = 300 m
y = 0 m
v₀ = v sin 30° = v/2
a = -9.8 m/s²
y = y₀ + v₀ t + ½ at²
0 = 300 + (v/2) t + ½ (-9.8) t²
0 = 300 + 1/2 v t − 4.9t²
Substitute:
0 = 300 + 1/2 (1200/√3) − 4.9t²
0 = 300 + 600/√3 − 4.9t²
4.9t² = 300 + 600/√3
t = 11.5 s
Solve for v:
v t = 1200/√3
v = 60.3 m/s
Graph:
desmos.com/calculator/1syounu0cg
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Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
Answer:
0.938 m/s.
Explanation:
Given:
ω = 1 rev in 0.67 s
In rad/s,
1 rev = 2pi rad
ω = 2pi ÷ 0.67
= 9.38 rad/s
Rp = 10 cm
= 0.1 m
V = ω × r
= 9.38 × 0.1
= 0.938 m/s.
Answer:
given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs
Explanation:
The Sun's gravitational pull keeps our planet orbiting the Sun <span>in a nice nearly-circular orbit.</span>
Force of gravity. Hope this is correct good luck!!
