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:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
, 
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
, 
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that 
, 
and 
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
There are 2 kilograms of water in 2 liters of water. if you convert the amount of water to kilograms it equals 2
To measure acceleration, a stopwatch is used.
