Answer:
dynamic and sometimes ballistic
Explanation:
(a) What is the maximum height the arrow will attain?
Given:
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
vᵧ = 0 m/s
aᵧ = -9.8 m/s²
Find: Δy
vᵧ² = v₀ᵧ² + 2aᵧΔy
(0 m/s)² = (24.5 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 30.7 m
(b) The target is at the height from which the arrow was shot. How far away is it?
Given:
Δy = 0 m
v₀ᵧ = 54 sin 27.0° m/s = 24.5 m/s
aᵧ = -9.8 m/s²
Find: t
Δy = v₀ᵧ t + ½ aᵧt²
0 m = (24.5 m/s) t + ½ (-9.8 m/s²) t²
t = 5.00 s
Given:
v₀ₓ = 54 cos 27.0° m/s = 48.1 m/s
aₓ = 0 m/s²
t = 5.00 s
Find: Δx
Δx = v₀ₓ t + ½ aₓt²
Δx = (48.1 m/s) (5.00 s) + ½ (0 m/s²) (5.00 s)²
Δx = 241 m
Answer:
466 km/h
255°
measured anticlockwise from the east direction
Explanation:
find the resultant velocity of a plane after a crosswind affects its motion. take the positive
x -direction as east, and the positive y -direction as north. The components of the plane's velocity (without wind) are
- <em>upx =</em> 0
- <em>upy = 450 Km/ h</em>
<em>And for the wind.</em>
- <em>uwx = - 120 Km/h</em>
- <em>uwy = 0</em>
<em>The components for the r</em><em>esultant velocity </em><em>of the plane are given by</em>
- <em>ux =upx + uwx = 0 - 120 Km/h = </em><u><em>- 120 Km/h</em></u>
- <em>uy = upy + uwy = - 450 Km/h +0 = </em><u><em>- 450 Km/h</em></u>
<em>The </em><em>magnitude </em><em>of the resultant velocity is</em>
<em />
= 466 km/h
Hope this helps u : )
Answer: he did travel 15 meters.
Explanation:
We have the data:
Acceleration = a = 1.2 m/s^2
Time lapes = 3 seconds
Initial speed = 3.2 m/s.
Then we start writing the acceleration:
a(t) = 1.2 m/s^2
now for the velocity, we integrate over time:
v(t) = (1.2 m/s^2)*t + v0
with v0 = 3.2 m/s
v(t) = (1.2 m/s^2)*t + 3.2 m/s
For the position, we integrate again.
p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0
Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m
Then the displacement at t = 3s will be equal to p(3s).
p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m
