How Do You Time a Projectile to Hit a Moving Train?

In summary, the problem involves calculating the time it takes for a bomb to drop from a balcony to hit a train passing below, as well as the time it takes for the train to move from one photogate to directly below the balcony. Measurements and accuracy will need to be determined in order to successfully drop the bomb, such as timing the train's movements and estimating its velocity. The equations x=(1/2)a*t^2+v(sub 0)+x (sub 0) and x=v(sub 0)*t+x(sub 0) can be used to calculate the time for the bomb to drop and the train to move, but the initial velocity of the bomb and the distance from point A will need to be determined as
  • #1
crystaluc12
3
0
Hi, I am a newbie in this forum, and I suck at physics, please help. :) Thank you

Homework Statement



You are standing on top floor anywhere (you choose) along a measuring tape (between the pts) that runs from point A to point B. There is a train on the ground level directly below that floor. The distance between the top floor and the level of the train is h=4.95 m. Train is going from Photogate A (a sensor that beeps when train goes through) (its directly below point A) to Photogate B (directly below point B). The train is running at a constant speed along the track and parallel to the balcony. Objective is to hit the train with a "bomb" blindly. "Bomb" will be thrown from balcony. You are given a stopwatch.

1. Calculate the time it takes your bomb to drop to the level of the train
2. Calculate time it takes the train to move form the first Photogate timer to directly below you.
3. What measurements will you need to make, estimate accuracy with which measurements need to be made, and how will you determine when to drop your projectile.

acceleration (a)=-g=-9.8 m/s^2


Homework Equations



I think:
x=(1/2)a*t^2+v(sub 0)+x (sub 0)
x=v(sub 0)*t+x(sub 0)

The Attempt at a Solution



1. For this one I thought that the initial velocity v(sub 0) would be equal to 0 since I would be dropping it from the top without any starting velocity. The initial height would be equal to 4.95 and acceleration is -9.8. So I did:

x=(1/2)a*t^2+v(sub 0)+x (sub 0)
0=(1/2)(-9.8)*t^2+0+4.95
-4.95=-4.9*t^2
t=square root (1.011)=approx 1

I am not sure if my calculations are correct.

2. I thought that it was calculated using x=v(sub 0)*t+x(sub 0)
so I thought that initial velocity of the train moving from a to b would be zero. So I put:
t=-x (sub 0)/v(sub 0)
But since v(sub 0) =0 the formula does not seem right to me. What formula do I need to use?

3. I thought that some measurements would be: I just drop my "bomb" many times from different measurements along the measuring tape and timed how long it took the "bomb" to drop and a) hit the train, b) miss the train, c) nearly hit the train. I need to know how long the train needs to get from 1st photogate to directly below me. I thought that listening for the beeps of the photogate and for sound of train and dropping it by a guess of when I thought the train was nearby would help me determine when to drop the "bomb"

I really am not sure if this is correct or if I am going at it the right way.

 
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  • #2
The problem states that the 'bomb' will be thrown, so the bomb will have any initial velocity. Also one can stand 'anywhere' do distance from point A is a variable, and one does not know the velocity of the train.

I think the problem is one of establish the distance from point A and velocity of the train as variables and relating them to the initial velocity of the bomb.

The equations:
x=(1/2)a*t^2+v(sub 0)+x (sub 0)
x=v(sub 0)*t+x(sub 0)

are correct keeping in mind that x is vertical position.

Ostensibly, one decides distance from A. Then how to determine the velocity of the train. Well knowing the length of the train, L, and the time it takes to pass gate A would give a speed. Then there is the time it takes to get to one's position, and there is the time interval the train spends passing one.
 
  • #3
Can someone please guide me in the right direction?

Dear newbie in the forum, first of all, welcome! Physics can be a challenging subject, but with practice and determination, you can excel at it. Let's take a look at your questions and see if we can provide some guidance.

1. Your approach to calculating the time it takes for the bomb to drop to the level of the train is correct. You are correct in assuming that the initial velocity is 0 since the bomb is simply being dropped from the top floor. However, your calculation is slightly off. The formula you used, x=(1/2)a*t^2+v(sub 0)+x (sub 0) is actually the formula for displacement, not time. The correct formula to use is x=v(sub 0)*t+(1/2)a*t^2. Plugging in the values, we have 4.95 = 0*t + (1/2)(-9.8)(t^2). Solving for t, we get t = √(4.95/-4.9) ≈ 1.01 seconds. So your calculation was close, but not quite right.

2. For this question, you are correct in assuming that the initial velocity of the train is 0, since it is starting from rest. However, the formula you used, x=v(sub 0)*t+x(sub 0) is not the correct formula to use in this situation. This formula is used for calculating displacement, not time. To calculate the time it takes for the train to reach the point directly below you, you can use the formula t = √(2x/a), where x is the distance between the first photogate and point A, and a is the acceleration due to gravity. Plugging in the values, we have t = √(2(4.95)/-9.8) ≈ 0.98 seconds.

3. Your approach to determining when to drop the "bomb" seems reasonable. However, you will need to be more precise in your measurements and timing in order to hit the train accurately. You can use a stopwatch to time the train as it moves from the first photogate to directly below you, and use that time to determine when to drop the "bomb." You can also use the sound of the train passing through the photogate as a cue to drop the "bomb." Additionally, you
 

FAQ: How Do You Time a Projectile to Hit a Moving Train?

What is velocity?

Velocity is a measure of the rate of change of an object's position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. The SI unit for velocity is meters per second (m/s).

How is velocity different from speed?

Velocity and speed are often used interchangeably, but they are actually different concepts. While speed is a measure of how fast an object is moving, velocity also takes into account the direction of the movement. This means that two objects can have the same speed but different velocities if they are moving in different directions.

What is acceleration?

Acceleration is a measure of the rate of change of an object's velocity with respect to time. It is also a vector quantity, and its direction is in the same direction as the change in velocity. The SI unit for acceleration is meters per second squared (m/s²).

How is acceleration related to velocity?

Acceleration and velocity are closely related. Acceleration is the change in velocity over time, meaning that it is the rate at which an object's velocity changes. If an object is accelerating, its velocity is changing, and if its acceleration is constant, its velocity will change at a steady rate.

What is the difference between average and instantaneous velocity/acceleration?

Average velocity and acceleration are calculated over a period of time, while instantaneous velocity and acceleration are measured at a specific moment in time. Average velocity is calculated by dividing the total displacement by the total time, while instantaneous velocity is the velocity at a specific point in time. Similarly, average acceleration is calculated by dividing the change in velocity by the change in time, while instantaneous acceleration is the acceleration at a specific point in time.

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