Answer:
15.67 m/s
Explanation:
The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.
Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by
where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:
Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:
Answer:
1. Can change the state of an object(rest to motion/ motion to rest)
2. May change the speed of an object if it is already moving.
3. May change the direction of motion of an object.
Explanation: A force acting on an object causes the object to change its shape or size, to start moving, to stop moving, to accelerate or decelerate.
Answer:
Part A
Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;
= -
Where;
= The action force
= The reaction force
Part B
The law indicates that the force with which a rocket ship uses in taking off from the Earth, is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-)
Part C
The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space
Explanation:
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.
Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).
At the start:
ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
