Solving a Vertical Throw Problem: Calculating Time, Speed, & Distance

In summary: Oh okay thanks for clearing that up, really appreciate it :D. It seems as if this user is not going to reply, but thanks for telling me about my mistake...thanks again :D.
  • #1
converse48
1
0

Homework Statement


A stone is thrown vertically upward with a speed of 13.5 m/s from the edge of a cliff 74.0 m high
(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

Homework Equations


i am only able to use these two equations
Δx=1/2at^2+Vo
Vo=initial velocity(i guess you guys already know that)

and
Vf(final velocity)=at+Vo

The Attempt at a Solution


for c, first thing i did was get 1.38s by using the final velocity equation
and i put that and all the other data into the delta x equation
Δx=1/2(-9.8)(1.38^2)+13.5(1.38)
Δx=-4.9(1.90)+13.5(1.38)
Δx=9.32m
so i added that to 74 to get 83.32
BUT ITS WRONG!
 
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  • #2
Okay, I'm assuming that you are using g = 9.8m/s² for your acceleration.

I would divide this problem into two parts, one for the time that it is going upward, then the other for the part when the stone is falling downwards.

For (a). We would have to find the time that it takes the stone to have a Vf of 0 because at that point it will start to fall. In other words...[tex]t=\frac{v_f - v_0}{a}[/tex] Since we know that for the first part t = -13.5/-9.8 = 1.38s (That is the time for it to go up and stop, then start falling down!) So JUST for the first part, it will take 2.76s for the stone to reach the start of the cliff again. Now we have to find out the second part and add the two up to get the total time. Now we have to use the second equation to get the rest of the time. We have to use the second equation because we don't know the [tex]v_{final}[/tex] when the stone hits the ground. So [tex]-74 = \frac{1}{2}at^2[/tex], and some playing around with the formula we get [tex]\frac{2*-74}{-9.8} = t^2[/tex] so taking the square root of that (to get the time) we get t = 3.89s. So adding all of the time up, we get [tex]t_{total} = 6.65s[/tex]
Note, I too am only in a class, so if there are please correct them :D.
 
  • #3
MysticDude said:
So [tex]-74 = \frac{1}{2}at^2[/tex], and some playing around with the formula we get [tex]\frac{2*-74}{-9.8} = t^2[/tex] so taking the square root of that (to get the time) we get t = 3.89s.

Close, but you're assuming that [itex]V_0 = 0[/tex] for the second part of the equation, which can't be: The stone reached it's peak and stopped at [itex]t \approx 1.38s[/tex], then accelerated downward from there, so it had to have some speed > 0 m/s when it once again reached the top of the cliff.
 
  • #4
zgozvrm said:
Close, but you're assuming that [itex]V_0 = 0[/tex] for the second part of the equation, which can't be: The stone reached it's peak and stopped at [itex]t \approx 1.38s[/tex], then accelerated downward from there, so it had to have some speed > 0 m/s when it once again reached the top of the cliff.

Oh okay thanks for clearing that up, really appreciate it :D. It seems as if this user is not going to reply, but thanks for telling me about my mistake...thanks again :D
 
  • #5


I would like to commend the student for their attempt at solving the problem using the given equations. However, there are a few issues with the solution provided.

Firstly, the equation Δx=1/2at^2+Vo is used to calculate the displacement (Δx) of an object undergoing constant acceleration (a) over a period of time (t) starting from an initial velocity (Vo). In this problem, we are not given the acceleration, so we cannot use this equation.

Secondly, the equation Vf=at+Vo is correct, but it is used to calculate the final velocity (Vf) of an object after a given time (t) starting from an initial velocity (Vo). In this problem, we are not asked to find the final velocity, so this equation is not needed.

To solve this problem, we can use the equation Δx=Vo*t+1/2at^2, where Vo is the initial velocity, t is the time, and a is the acceleration due to gravity (which is -9.8 m/s^2 in this case).

(a) To find the time it takes for the stone to reach the bottom of the cliff, we can rearrange the equation to t=√(2Δx/a). Plugging in the values, we get t=√(2*74/9.8)≈3.05 seconds.

(b) To find the speed just before hitting the ground, we can use the equation Vf=Vo+at. As the stone is just about to hit the ground, its final velocity (Vf) will be 0. So, we can rearrange the equation to Vo=-at. Plugging in the values, we get Vo=-9.8*3.05≈-29.89 m/s. This is the magnitude of the velocity, so the actual speed will be 29.89 m/s.

(c) To find the total distance traveled by the stone, we can use the equation Δx=Vo*t+1/2at^2. Plugging in the values, we get Δx=13.5*3.05+1/2*(-9.8)*(3.05)^2≈83.26 meters. This is the total distance traveled by the stone, which is equal to the height of the cliff (74 meters) plus the additional distance traveled while
 

Related to Solving a Vertical Throw Problem: Calculating Time, Speed, & Distance

What is a vertical throw problem?

A vertical throw problem is a physics problem that involves calculating the time, speed, and distance of an object that is thrown vertically into the air. It typically involves the use of equations such as the kinematic equations and the equation for projectile motion.

What information do I need to solve a vertical throw problem?

To solve a vertical throw problem, you will need to know the initial velocity of the object, the acceleration due to gravity, and the final position of the object. You may also need to know the angle at which the object was thrown, if it was thrown at an angle rather than straight up.

How do I calculate the time, speed, and distance of a vertical throw?

The time, speed, and distance of a vertical throw can be calculated using the kinematic equations. The equations are:

  • Time (t): t = (vf - vi) / a
  • Speed (vf): vf = vi + at
  • Distance (d): d = vit + 1/2at2

Where vf is the final velocity, vi is the initial velocity, a is the acceleration due to gravity, and t is the time.

What units should I use when solving a vertical throw problem?

When solving a vertical throw problem, it is important to use consistent units throughout your calculations. The most commonly used units for time, speed, and distance are seconds (s), meters per second (m/s), and meters (m), respectively. However, you can use other units as long as they are consistent throughout your calculations.

What are some common mistakes when solving a vertical throw problem?

Some common mistakes when solving a vertical throw problem include using the wrong units, not paying attention to the direction of the velocity and acceleration, and not taking into account the effects of air resistance. It is also important to double check your calculations and pay attention to significant figures to ensure accurate results.

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