Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
'A' is the the point on the graph that shows a temperature of 40°C and the time of 25 minutes
If the car's motion appears as a horizontal line on a <u><em>position-time </em></u>graph, it shows that as time changes, the car's position doesn't change.
This is just a complicated way to say that the car is <em>not moving</em>.<em> (A)</em>
Distance from the reference point
The final position of the object after 2 s is 11 m.
Motion: This can be defined as the change in position of a body.
⇒ Formula:
- x = x₀+v₀t+1/2(at²)........................ Equation 1
⇒ Where:
- x = Final position of the object
- x₀ = Starting position
- v₀ = Starting velocity
- t = time
- a = acceleration
From the question,
⇒ Given:
- x₀ = 4.5 m/s
- t = 2 s
- x₀ = 2m
- a = 0 m/s²
⇒ Substitute these values into equation 1
- x = 2+(4.5×2)+1/2(0²×2)
- x = 2+9+0
- x = 11 m
Hence, The final position of the object after 2 s is 11 m
Learn more about motion here: brainly.com/question/15531840
