1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s
2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m
Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.
For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.
If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.
Answer:
20 N/m
Explanation:
From the question,
The ball-point pen obays hook's law.
From hook's law,
F = ke............................ Equation 1
Where F = Force, k = spring constant, e = compression.
Make k the subject of the equation
k = F/e........................ Equation 2
Given: F = 0.1 N, e = 0.005 m.
Substitute these values into equation 2
k = 0.1/0.005
k = 20 N/m.
Hence the spring constant of the tiny spring is 20 N/m
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.
Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is
(487-left + 632-right) - (632-right - 487-right) = 145N to the right.
The net force on the car is all to the right.
The car accelerates to the right.
here we will use the concept of Newton's III law
as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat
So here we will say
impulse gain by the ball = impulse lost by the bat
given that
For ball the change in speed will be
now from above equation
so speed of bat will decrease by 6.72 mph