Constant Acceleration Question

The maximum height should be reached after 3.6 seconds, not 2 seconds. This means that the calculation for the height of the tower should be revised to:x-xo=19.6*3.6+.5*-9.8*3.6^2x-xo=66.816 mIn summary, the rock is thrown vertically upward from ground level with an initial velocity of 19.6 m/s. After 3.6 seconds, it reaches its maximum height of 66.816 m, passing the top of a tall tower at 1.6 seconds. The discrepancy in the calculation was due to using 2 seconds instead of 3.6 seconds as the time for the rock to
  • #1
phyphyphy
3
0
A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 2.0 s later it reaches its maximum height. What is the height of the tower?

My thought process:

First find initial velocity:

v=vo+at
0=vo+(-9.8)(2)
vo=19.6 m/s

Then you can find the height of the tower, where x-xo is the displacement/height of the tower:
x-xo=vo*t+.5*a*t^2
x-xo=19.6*1.6+.5*-9.8*1.6^2
x-xo=18.816 m

However, WileyPlus says that I am wrong. What am I doing wrong?
 
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  • #2
phyphyphy said:
0=vo+(-9.8)(2)
vo=19.6 m/s

You did almost everything correctly. But is it true that it's maximum height is reached after 2 seconds?
(That's what your math says.)
 
  • #3
OH, so I should have the maximum height as 3.6 seconds, correct?
 
  • #4
phyphyphy said:
OH, so I should have the maximum height as 3.6 seconds, correct?

Correct.
 
  • #5


There could be a few things that could be causing your answer to differ from WileyPlus's answer. Here are some things you can check to make sure your calculation is correct:

1. Make sure you are using the correct units for all variables. In this problem, the acceleration is given in m/s^2, so make sure your velocity (vo) is also in m/s and your time (t) is in seconds.

2. Check your calculations for any errors. Make sure you are using the correct values and that your calculations are correct. It's always a good idea to double check your work to ensure accuracy.

3. Make sure you are using the correct formula. In this problem, you are looking for the height of the tower, so you should be using the formula x-xo=vo*t+.5*a*t^2, where xo is the initial position (in this case, ground level).

4. Consider the direction of motion. Since the rock is thrown vertically upward, the acceleration due to gravity should be negative. Make sure you are using the negative sign in your calculations.

5. Check your answer against a different source. If you are still unsure of your answer, try using a different resource or asking a classmate or teacher for help. Sometimes, different sources may use slightly different values or formulas, so it's always a good idea to double check your answer using a different source.
 

1. What is constant acceleration?

Constant acceleration is the rate at which an object's velocity changes over time remains constant. In other words, it means that the object's speed is increasing or decreasing by the same amount every second.

2. How is constant acceleration different from uniform motion?

Uniform motion is when an object moves with a constant speed in a straight line, while constant acceleration means the object's speed is changing by the same amount every second. In other words, constant acceleration involves a change in speed, while uniform motion does not.

3. What is the equation for calculating constant acceleration?

The equation for constant acceleration is a = (v - u) / t, where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time taken.

4. How is constant acceleration represented on a graph?

On a velocity vs. time graph, constant acceleration will appear as a straight, diagonal line. The slope of the line represents the acceleration, and the area under the line represents the distance traveled by the object.

5. What are some real-life examples of constant acceleration?

Some examples of constant acceleration in everyday life include a car accelerating from a stop, a rollercoaster going down a hill, a skydiver falling towards the ground, or a rocket launching into space.

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