# Projectile physics homework problem

• UNknown 2010
In summary,You have solved the problem of projectile motion, but your answer does not match the answer in the book. Your answer is 32 m/s, while the answer in the book is 37 m/s. You might want to check your numbers again.
UNknown 2010
Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?

Thanks =)

Last edited by a moderator:

Maybe you will see my mistake . . .

I did not trust that v² formula because the signs are complicated: you would get the 32.08 answer regardless of whether the initial velocity was up or down! So for the vertical part I used d = vt+.5gt² with d = -50, v = +7sin(53), g= -9.81 to get that the time to fall is 3.37 s. Then I put that into
Vf = Vi + at = 7sin(53)-9.81*3.37 = -27.5 m/s.

UNknown 2010 said:
Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?Thanks =)
I believe the answer you gave above (your answer) is correct, with possible, minor precision issues. The mistake must be in the book.

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i got exactly your answer Unknown 2010, so the book is wrong

and Delphi51 you don't even need to look at time

^^same thing with me...the book seems to be wrong..

Delphi51 said:
So for the vertical part I used d = vt+.5gt² with d = -50, v = +7sin(53), g= -9.81 to get that the time to fall is 3.37 s.
You might want to check your numbers again. I got t = 3.813 seconds using your method.

I still don't see it, Collinsmark. Hope you will pinpoint it for me!
-50 = 7*sin(53) - .5*9.81*t²
4.905*t² = 7*sin(53) + 50
4.905*t² = 55.59
t² = 55.59/4.905 = 11.33
t = 3.367

edit - Oops: forgot the t! Now getting the 31.8 m/s for the vertical.
Thank you for sorting that out!

Last edited:

Delphi51 said:
I still don't see it, Collinsmark. Hope you will pinpoint it for me!
-50 = 7*sin(53) - .5*9.81*t²
4.905*t² = 7*sin(53) + 50
4.905*t² = 55.59
t² = 55.59/4.905 = 11.33
t = 3.367

edit - Oops: forgot the t! Now getting the 31.8 m/s for the vertical.
Thank you for sorting that out!

i honestly don't see how your solution is plausible of working. Can you explain how you take into account the time traveled above 50m?

The t is the time of flight. Those formulas Vf = Vi + at and
d = Vi*t + ½a*t² are for all constant acceleration motion and always account for the time, distance and speed automatically whether above or below the starting point. If you solve the first formula for t and sub into the second, you'll get the Vf² = Vi² + 2ad formula.

UNknown 2010 said:
Hello,

I have solved the following problem but my answer does not matched with the answer which is written in the book. My answer is 32 m/s but the answer of the book is 37 m/s

[PLAIN]http://img842.imageshack.us/img842/6706/projectile.png

Could anyone please show me where my mistake is ?

Thanks =)

Your work is perfect, with the minor exception that $$v_y$$ should be negative since the rock is moving downward at that point. This will also affect your angle. Remember that $$\sqrt{x^2} = \pm x$$

Last edited by a moderator:

Delphi51 said:
The t is the time of flight. Those formulas Vf = Vi + at and
d = Vi*t + ½a*t² are for all constant acceleration motion and always account for the time, distance and speed automatically whether above or below the starting point. If you solve the first formula for t and sub into the second, you'll get the Vf² = Vi² + 2ad formula.

okay i see what you did, and why it works

## 1. What is a projectile?

A projectile is any object that is thrown or launched into the air and is only affected by gravity and air resistance. Examples include a ball thrown in the air or a bullet shot from a gun.

## 2. How do you calculate the initial velocity of a projectile?

The initial velocity of a projectile can be calculated using the equation V0 = V * cos(θ), where V is the initial velocity magnitude and θ is the angle of launch.

## 3. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the equation Hmax = (V02 * sin2(θ)) / (2 * g), where V0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation R = (V02 * sin(2θ)) / g, where V0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 5. How do you take air resistance into account in projectile motion?

In most cases, air resistance can be ignored when calculating projectile motion. However, if it needs to be taken into account, it can be incorporated using the drag force equation, which is FD = 1/2 * ρ * CD * A * V02, where ρ is the air density, CD is the drag coefficient, A is the cross-sectional area of the object, and V0 is the initial velocity.

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