# Projectile Motion -- position and the highest point

• Masrat_A
Solution for (b):## x = 40 t = 200####t = 5## sec##y = 30(5) - 5(5)^2 = -25 m####x = 40(5) = 200 m####(200, -25)##For #4:How much time does it take to fall 20 m ?To find the time it takes for an object to fall a certain distance, we can use the equation ##h = \frac{1}{2}gt^2##, where h is the height, g is the
Masrat_A
For 3 (a), I feel like I've gotten it down correctly, but comments would be appreciated! 3 (b), I'm not too sure about; could anyone please give a thorough review and point out mistakes if there are any?

Number 4, I have no idea where to even begin. I've spent nearly an hour brainstorming, but have gotten nowhere. I do have class notes, but I'm unable to make sense of any of it. May I please have some direction on how I could go about solving? Are there any formulas I should be looking into?

1. Homework Statement

3) A projectile is fired from ground at 37 degrees above horizontal at speed 50 m/s.
a) Determine the position of the highest point and the time it takes to reach it.
b) Determine the position and time at which it strikes a wall 200 m away.

4) (See figure.) A lemming accelerates uniformly starting from rest 2 m before jumping horizontally from the top of a 20 m high cliff. What is the minimum acceleration required so that the lemming can clear the rocks stretching out 4 m from the base of the cliff?

## Homework Equations

Figure for question 4: http://i.imgur.com/SMspyLdh.jpg

## The Attempt at a Solution

http://i.imgur.com/MpjwWeih.jpg

I cannot make sense of some things in your solution. I'd recommend using Latex - there are few equations it is not difficult or tidy things up a little bit.

I tend to work these problems in general terms, so that we derive formulas that can be applied to similar problems in the future.

For problem 3, I would begin with:

##a_y=-g##

##a_x=0##

Resolving the initial velocity into ##xy##-components, we have:

##v_{y_0}=v_0\sin(\theta)##

##v_{x_0}=v_0\cos(\theta)##

Now, using the fact that acceleration is the time rate of change of velocity, can you state the initial value problems (IVPs) whose solutions are the ##xy##-components of the projectile's velocity and solve the IVPs?

Masrat_A said:
Are there any formulas I should be looking into?
Equation of trajectory of a projectile in 2D.

For part 3(b),if you are familiar with the equation of trajectory of a projectile,that would very easily give you the position of the required point.

The time is pretty easy to calculate,since we know the horizontal component of the velocity(vcostheta) and distance travelled(200m)..then it's simply v=d/t

Question 4 can be broken into two parts..one projectile and one 1D kinematics

Once again,if you know what the equation of trajectory of a projectile is and how to apply it,the velocity will be quite easy to find.
Then comes the 1D kinematics part
Say the velocity you found out is v.
Your new question becomes "What constant acceleration should a lemming starting from rest have so that it can attain a velocity v within 2m??"
That should again be simple ,using Newton's equations of motion.

QuantumQuest said:
I cannot make sense of some things in your solution. I'd recommend using Latex - there are few equations it is not difficult or tidy things up a little bit.

Here are my solutions to problem 3 (a) and (b) using latex.

##v_ox = 50\cos37^\circ = 40 = v_x##
##v_oy = 50\cos37^\circ = 30##
##x = 40t##

## v_y = 30 - 10t##
##y = 30t - t5^2##
##v_y^2 = 900 - 20y##

Solution for (a):

##v_y = 0##
##30 - 10t = 0##
##t = 3## sec

##y = 20(3) - 5(3)^2 = 15 m##
##x = 40(3) = 120 m##
##(120, 15)##

Solution for (b):

## x = 40 t = 200##
##t = 5## sec

##y = 20(5) - 5(5)^2 = -25##
##x = 40(5) = 200##
##(200, -25)##

Tanishq Nandan said:
Equation of trajectory of a projectile in 2D.

For part 3(b),if you are familiar with the equation of trajectory of a projectile,that would very easily give you the position of the required point.

The time is pretty easy to calculate,since we know the horizontal component of the velocity(vcostheta) and distance travelled(200m)..then it's simply v=d/t

Question 4 can be broken into two parts..one projectile and one 1D kinematics

Once again,if you know what the equation of trajectory of a projectile is and how to apply it,the velocity will be quite easy to find.
Then comes the 1D kinematics part
Say the velocity you found out is v.
Your new question becomes "What constant acceleration should a lemming starting from rest have so that it can attain a velocity v within 2m??"
That should again be simple ,using Newton's equations of motion.

At this moment, unfortunately, I am not very much familiar with equation of trajectory of a projectile. Is there any other method by which I could find the position for question 3b and velocity for question 4?

In addition, for 3b, would you please mind explaining which would be the horizontal component of velocity?

Last edited by a moderator:
That looks like the correct method, but you have a typo and are using an incorrect value as indicated below.

(You must be using 10 m/s2 as a value for g.)
Masrat_A said:
Here are my solutions to problem 3 (a) and (b) using latex.
##v_ox = 50\cos37^\circ = 40 = v_x##
##v_oy = 50\cos37^\circ = 30##
##x = 40t##

## v_y = 30 - 10t##
##y = 30t - t5^2##
You should have squared t, not g/2 .
##v_y^2 = 900 - 20y##

Solution for (a):

##v_y = 0##
##30 - 10t = 0##
##t = 3## sec

##y = 20(3) - 5(3)^2 = 15 m##
You appear to have used a value of 20 m/s for ##\ (v_0)_y \ ## rather than 30 m/s.
You also did this for part (b).
##x = 40(3) = 120 m##
##(120, 15)##

Solution for (b):

## x = 40 t = 200##
##t = 5## sec

##y = 20(5) - 5(5)^2 = -25##
##x = 40(5) = 200##
##(200, -25)##
For #4:
How much time does it take to fall 20 m ?

SammyS said:
That looks like the correct method, but you have a typo and are using an incorrect value as indicated below.

(You must be using 10 m/s2 as a value for g.)
You should have squared t, not g/2 .You appear to have used a value of 20 m/s for ##\ (v_0)_y \ ## rather than 30 m/s.
You also did this for part (b).

Thank you for pointing those out! Here are my updated solutions; do these look better?

##v_ox = 50\cos37^\circ = 40 = v_x##
##v_oy = 50\cos37^\circ = 30##
##x = 40t##

## v_y = 30 - 10t##
##y = 30t - 5t^2##
##v_y^2 = 900 - 20y##

Solution for (a):

##v_y = 0##
##30 - 10t = 0##
##t = 3## sec

##y = 30(3) - 5(3)^2 = 45 m##
##x = 40(3) = 120 m##
##(120, 45)##

Solution for (b):

## x = 40 t = 200##
##t = 5## sec

##y = 30(5) - 5(5)^2 = 25##
##x = 40(5) = 200##
##(200, 25)##

SammyS said:
For #4:
How much time does it take to fall 20 m ?

Assuming that the initial velocity is 0, this is what I have been able to come up with for time:

##d = V_i * t + \frac{1}2at^2##
##20 = \frac{1}29.8t^2##
##t = \sqrt{\frac{20}{4.9}}##
##t = 2.02##

Minimum acceleration:

##d = vt##
##4 = v * 2.02##
##v = \frac{4}{2.02}##
##v = 1.98 m/s##

##v_f^2 = 3.92##
##3.92 = 2 * ax = 4 * a##
##a = \frac{3.92}{4}##
##a = 0.98 m/s^2##

Masrat_A said:
##v_oy = 50\cos37^\circ = 30##

Be careful here ##v_oy = 50\sin37^\circ##

Now, ##y = v_0\sin(\theta) t - \frac{1}{2}gt^2## as you wrote it (with values substituted) but as SammyS points out you must square ##t## not ##\frac{g}{2}## . We usually take ##g = 9.8 \frac{m}{s^2}## unless specified otherwise. Also, as mentioned, you must put ##30 \frac{m}{s}## in the equation of ##y## for the initial velocity component ##v_0y##. Also, for part (b) the time you find is correct but again ##v_0y## must be corrected.

Masrat_A said:
Thank you for pointing those out! Here are my updated solutions; do these look better?

##v_ox = 50\cos37^\circ = 40 = v_x##
##v_oy = 50\cos37^\circ = 30##
##x = 40t##

## v_y = 30 - 10t##
##y = 30t - 5t^2##
##v_y^2 = 900 - 20y##

Solution for (a):

##v_y = 0##
##30 - 10t = 0##
##t = 3## sec

##y = 30(3) - 5(3)^2 = 45 m##
##x = 40(3) = 120 m##
##(120, 45)##

Solution for (b):

## x = 40 t = 200##
##t = 5## sec

##y = 30(5) - 5(5)^2 = 25##
##x = 40(5) = 200##
##(200, 25)##
Assuming that the initial velocity is 0, this is what I have been able to come up with for time:

##d = V_i * t + \frac{1}2at^2##
##20 = \frac{1}29.8t^2##
##t = \sqrt{\frac{20}{4.9}}##
##t = 2.02##

Minimum acceleration:

##d = vt##
##4 = v * 2.02##
##v = \frac{4}{2.02}##
##v = 1.98 m/s##

##v_f^2 = 3.92##
##3.92 = 2 * ax = 4 * a##
##a = \frac{3.92}{4}##
##a = 0.98 m/s^2##
That looks good for both problems.

It's interesting that you have switched to using a value for g of 9.8 m/s2 for problem 4.

## 1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched or thrown into the air and moves under the influence of gravity. This type of motion follows a curved path, known as a projectile trajectory.

## 2. How is the position of a projectile determined during its motion?

The position of a projectile is determined by its horizontal and vertical components of motion. The horizontal position can be calculated using the formula x = x0 + v0x*t, where x0 is the initial position, v0x is the initial velocity in the x-direction, and t is the time. The vertical position can be calculated using the formula y = y0 + v0y*t - 1/2*g*t^2, where y0 is the initial position, v0y is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time.

## 3. What is the highest point in a projectile's trajectory?

The highest point in a projectile's trajectory is known as the maximum height or peak. This is the point where the projectile's vertical velocity becomes zero before it starts to fall back down due to gravity.

## 4. How can the highest point in a projectile's trajectory be calculated?

The highest point in a projectile's trajectory can be calculated using the formula y = y0 + (v0y^2/2g), where y0 is the initial position, v0y is the initial velocity in the y-direction, and g is the acceleration due to gravity. This formula gives the maximum height or peak of the projectile's trajectory.

## 5. Can the highest point of a projectile's trajectory be affected by air resistance?

Yes, air resistance can affect the highest point of a projectile's trajectory. In the presence of air resistance, the projectile will experience a smaller vertical velocity and will not reach the same maximum height as it would without air resistance. This can be taken into account by adjusting the initial velocity and acceleration values in the calculation of the highest point.

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