C to raise or lower lightweight objects
As this is the projectile motion means the motion of object under constant acceleration, so we use all eq. of motion under constant acceleration.
Angle above horizontal=θ= 35°
Initial speed= v₁ = 25 m/s
Time of flight= t= 2.55 s
Now,
from eq. of motion
x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²
As, there is no acceleration in horizontal direction so a=0 and also initial displacement x₁=0
x= 25*cos(35)*2.55
x=52.22 m
Answer:
2 m/s
Explanation:
Applying,
The law of conservation of momentum
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')............... Equation 1
Where m = mass of the first freight car, m' = mass of the second freight car, u = initial velocity of the first freight car, u' = initial velocity of the second freight car, V = final combined velocity/ speed.
make V the subject of the equation
V = (mu+m'u')/(m+m')........... Equation 2
From the question,
Given: m = 1234 kg, m' = 2468 kg, u = 6 m/s, u' = 0 m/s (at rest)
Substitute these values into equation 2
V = [(1234×6)+(2468×0)]/(1234+2468)
V = 7404/3702
V = 2 m/s
Answer:
a) y = 16.51 [m]
b) t = 1.83 [s]
Explanation:
To solve this problem we must use two kinematics equations, the first to determine the height to which the ball reaches, and the second equation to determine how long it lasts in the air.
where:
Vf = final velocity = 0
Vi = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81[m/s^2]
t = time [s]
Note: the negative sign of the Equation indicates that the acceleration of gravity acts in the opposite direction to the movement of the ball. The final velocity is zero, since the ball reaches its maximum altitude when the velocity is zero.
Now replacing:
0 = (18)^2 - (2*9.81*y)
y = 16.51 [m]
b)
0 = 18 - (9.81*t)
t = 1.83 [s]
