A) The vertical component of velocity v is taking the rock to a height
Vertical component =
The time taken to reach maximum height =
So total time of rocks flight =
Range of rock is due to the horizontal component of velocity =
Range =
=
Maximum height =
=
Since range = maximum height
We have
So when angle of projection is
range is equal to maximum height reached.
b) We have range =
=
Maximum of range is reached when
Maximum range =
c) For range to be equal to maximum height only condition is
, it does not depend upon acceleration due to gravity and velocity. That angle is a constant.
Answer:
V = 10m/s for the final velocity.
S = 10m for distance II.
Explanation:
Given the given information:
5m/s2 acceleration
2 seconds Equals time
Because the vehicle begins at rest, its initial velocity is 0m/s.
The first equation of motion would be used to get the ultimate velocity.
V = U + at
We have by substituting into the equation
V = 0 + 5*2
V = 0 + 10
V = 10m/s.
Now, to find the distance covered, we would use the second equation of motion.
S = ut + ½at²
S = 0*2 + ½*5*2²
S = ½*5*4
S = 20/2
S = 10m
Explanation:
the work a force does is measured in newton-meters (N-m), which use the symbol of J. Time (t) is expressed in Second. Power (P) is expressed in Watt (W).
Answer:
Part a)
at t = 3.00 s
at t = 20.0 s
Part b)
at t = 3.00 s
at t = 20.0 s
Explanation:
The car starts at x = 0
Part a)
Now at t = 3.00 s
the position of the car is given as x = 25 m and its speed is given as v = 11 m/s
Now for average velocity we have
Now for average acceleration we have
Part b)
Now at t = 20.0 s
the position of the car is given as x = 385 m and its speed is given as v = 45 m/s
Now for average velocity we have
Now for average acceleration we have