A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?
1 answer:
Answer:
C. 30.6m
Explanation:
To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:
Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.
Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)
You might be interested in
Answer:
the mass drop by 6.5cm before coming to rest.
Explanation:
Given that:
the mass of the block M= 1.40 kg
angle of inclination θ = 30°
spring constant K = 40.0 N/m
mass of the suspended block m = 60.0 g = 0.06 kg
initial downward speed = 1.40 m/s
The objective is to determine how far does it drop before coming to rest?
Let assume it drops at y from a certain point in the vertical direction;
Then :
Workdone by gravity on the mass of the block is:
= - 6.867y
Workdone by gravity on the mass of the suspended block is:
= 0.5886
The workdone by the spring = -1/2ky²
= -0.5 × 40y²
= -20 y²
The net workdone is = -20 y² - 6.867y + 0.5886y
According to the work energy theorem
Net work done = Δ K.E
-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²
-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176
-20 y² - 6.867y + 0.5886 = 0.0588
-20 y² - 6.867y + 0.5886 - 0.0588
-20 y² - 6.867y + 0.5298 = 0
multiplying through by (-)
20 y² + 6.867y - 0.5298 = 0
Using the quadratic formula:
where;
a = 20 ; b = 6.867 c= - 0.5298
= 0.0649 OR −0.408
We go by the positive integer
y = 0.0649 m
y = 6.5 cm
Therefore; the mass drop by 6.5cm before coming to rest.