A ball is shot up from the ground with an initial vertical velocity of 30 m/s. What is the velocity of the ball as it reaches it
1 answer:
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A) The position at t = 2.0 sec is 43.0 m east
B) The position is 55 m east
Explanation:
A)
In order to solve the problem, we take the east direction as positive direction.
We know that:
- at t = 0, the motorcyclist is at a position of
- at t = 0, the initial velocity of the motorcyclist is east
- The acceleration of the motorcyclist is constant and it is
Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression
where t is the time.
Substituting t = 2.0 s, we find the position:
B)
The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:
where:
is the initial velocity
is the acceleration
We want to find the time t at which the velocity is
v = 25 m/s
Solving the equation for t,
And therefore, the position at t = 2.5 s is:
Learn more about accelerated motion:
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