An experiment to measure the value of g is constructed using a tall tower outfit

Question

An experiment to measure the value of g is constructed using a tall tower outfitted with two sensing devices, one a distance H above the other. A small ball is fired straight up in the tower so that it rises to near the top and then falls back down; each sensing device reads out the time that elapses between the ball going up past the sensor and back down past the sensor.

(a) It takes a time 2t1 for the ball to rise past and then come back down past the lower sensor, and a time 2t2 for the ball to rise past and then come back down past the upper sensor. Find an expression for g using these times and the height H. (Use the following as necessary: t1, t2, and H.)

(b) Determine the value of g if H equals 21 m, t1 equals 4 s, and t2 equals 3.42 s.

Thank you for the help- I can't figure out how to set up the first part and get an equation that will actually allow me to solve the second part. Let me know what steps you used. Thanks!

Explanation / Answer

So we have the following data with us

Time recorded by the lower sensor, 2t1

Time recorded by the upper sensor, 2t2

Distance between the two sensors, H

We need to find an expression for g, the acceleration due to gravity.

(a)

Let us take

The top of the building as point A

The upper sensor as point B

The lower sensor as point C

AB = h2, AC = h1, BC = H

So, we have,

AC - AB = h1 - h2 = BC = H ----- (Eq.1)

To calculate the values of AC and AB, we need the kinematic equations.

Since the value of acceleration is constant irrespective of height, we can use the kinematic equations of motion.

From the second kinematic equation, we have

h = ut + (1/2)*g*t2

where, h = height, u = initial velocity and t = time taken

We also know that the time taken by the ball to ascend to the top of the tower from a point would be equal to the time taken by the ball to descend to that same point from the top of the tower i.e. C to A will take t1 and A to C will take t1 as well. (You can prove it using the same kinematic equations of motion, I can explain this part further if required)

So we can half the times recorded by the sensors and use them to calculate h1 and h2

Since the initial velocity of the ball at the top of the tower is zero, u = 0

So,

h1 = (1/2)*g*t12 ------- (Eq.2)

h2 = (1/2)*g*t22 ------- (Eq.3)

Using Eq.2 and Eq.3 in Eq.1, we get

h1 - h2 = (1/2)*g*(t12 - t22) = H

So,

g = 2H / (t12 - t22)

(b) H = 21 m

t1 = 4 s

t2 = 3.42 s

Using the above values in our equation, we have

g = 2 * 21 / (42 - 3.422) = 9.759 m/s2

Ans. g = 9.76 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.