A mouse jumps horizontally from a box of height 0.25m. If the mouse jumps with a speed of 2.1 m/s, how far from the box does th
2 answers:
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Answer:
the distance in meters traveled by a point outside the rim is 157.1 m
Explanation:
Given;
radius of the disk, r = 50 cm = 0.5 m
angular speed of the disk, ω = 100 rpm
time of motion, t = 30 s
The distance in meters traveled by a point outside the rim is calculated as follows;
Therefore, the distance in meters traveled by a point outside the rim is 157.1 m
1a) Bill and the dog must have a speed of 13.0 m/s
1b) The speed of the dog must be 22.5 m/s
2a) The ball passes over the outfielder's head at 3.33 s
2b) The ball passes 1.2 m above the glove
2c) The player can jump after 2.10 s or 3.13 s after the ball has been hit
2d) One solution is when the player is jumping up, the other solution is when the player is falling down
Explanation:
1a)
The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
In part a), we want to know at what speed Bill and the dog have to run in order to intercept the ball as it lands on the ground: this means that Bill and the dog must have the same velocity as the horizontal velocity of the ball.
The ball's initial speed is
u = 15 m/s
And the angle of projection is
So, the ball's horizontal velocity is
And therefore, Bill and the dog must have this speed.
1b)
For this part, we have to consider the vertical motion of the ball first.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
is the acceleration of gravity
The ball is at a position of y = 2 m above the ground when:
Which has two solutions: t=0.34 s and t=1.19 s. We are told that the ball is falling to the ground, so we have to consider the second solution, t = 1.19 s.
The horizontal distance covered by the ball during this time is
The dog must be there 0.5 s before, so at a time
t' = t - 0.5 = 0.69 s
So, the speed of the dog must be
2a)
Here we just need to consider the horizontal motion of the ball.
The horizontal distance covered is
while the horizontal velocity of the ball is
where u = 34 m/s is the initial speed.
So, the time taken for the ball to cover this distance is
2b)
Here we need to calculate the vertical position of the ball at t = 3.33 s.
The vertical position is given by
where
h = 1.2 m is the initial height
is the initial vertical velocity
is the acceleration of gravity
Substituting t = 3.33 s,
And sinc the glove is at a height of y' = 2.3 m, the difference in height is
y - y' = 3.5 - 2.3 = 1.2 m
2c)
In order to intercept the ball, he jumps upward at a vertical speed of
So its position of the glove at time t' is
where h' = 2.3 m is the initial height of the glove, and t' is the time from the moment when he jumps. To catch the ball, the height must be
y' = y = 3.5 m (the height of the ball)
Substituting and solving for t', we find
Which has two solutions: t' = 0.20 s, t' = 1.23 s. But this is the time t' that the player takes to reach the same height of the ball: so the corresponding time after the ball has been hit is
So we have two solutions:
So, the player can jump after 2.10 s or after 3.13 s.
2d)
The reason for the two solutions is the following: the motion of the player is a free fall motion, so initially he jump upwards, then because of gravity he is accelerated downward, and therefore eventually he reaches a maximum height and then he falls down.
Therefore, the two solutions corresponds to the two different part of the motion.
The first solution, t'' = 2.10 s, is the time at which the player catches the ball while he is in motion upward.
On the other hand, the second solution t'' = 3.13 s, is the time at which the player catches the ball while falling down.
Learn more about projectile motion:
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