<span>Aristotle would say that an object comes to rest because it got tired. Aristotle came up with ideas, but never tested them.
Galileo used experiments and mathematical models to analyze motion. He used a slanted inclined plane an order to slow down the acceleration due to gravity on an object. He found out that an object accelerates as it falls. </span>
Answer:
a) 103.32 m
b) 9.18 s
Explanation:
a) Let's use the knowledge that at the top of its trajectory, the baseball will have a final velocity of 0 m/s.
The acceleration due to gravity is -9.8 m/s², assuming the upwards direction is positive and the downwards direction is negative.
The initial velocity of the baseball is 45 m/s.
We are trying to find the vertical displacement of the baseball, Δx, and we have the variables v, a, and v₀.
Find the constant acceleration equation that contains all four of these variables:
Substitute the known values into the equation.
- (0)² = (45)² + 2(-9.8)Δx
- 0 = 2025 - 19.6Δx
- -2025 = -19.6Δx
- Δx = 103.32
The maximum height of the ball before it falls back down is 103.32 m.
b) Now we want to solve for time t. Find a constant acceleration equation that contains three known variables.
Substitute known values into this equation.
- 0 = 45 + (-9.8)t
- -45 = -9.8t
- t = 4.59183673
Remember that this is only half of the baseball's flight since we are using the final velocity for when the ball is halfway through its trajectory.
To solve for the total time the baseball is in the air, double the time t we solved for.
The baseball is in the air for 9.18 s.
Answer:
C. The floor pushing back against the foot
Explanation:
Answer:
F t = m Δv impulse delivered = change in momentum
Δv = 100 * .1 / .5 = 20 m/s original speed of puck
KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck
E = μ m g d energy lost by puck
Ff = μ m g = m a deceleration of puck due to friction
a = μ g = 9.8 * .2 = 1.96 m/s^2
v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box
m v2 = M V conservation of momentum when puck strikes box
V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision
KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision
KE = μ M g d energy lost by box in sliding distance d
d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides
Answer:
Part a)
Part b)
Explanation:
Part a)
In order to have same range for same initial speed we can say
so after comparing above we will have
so we have
Part b)
Time of flight for the first ball is given as
Now for other angle of projection time is given as
So here the time lag between two is given as
