Homework Help: What was the bullet's speed as it left the barrel?

1. Sep 28, 2007

AvrGang

I'm a freshman student in the university, and I have to solve 50 questions to finish my assignment on the projectile motion, and I'm stuck with these three questions. I wish anybody can help me as fast as he or she can, because it's my first time to study the projectile motion. (My assignment is due to sunday 30th september).

1. The problem statement, all variables and given/known data

Here is the first problem: A refile is aimed horizontally at a target 50 m away. The bullet hits the target 2.0 cm below the aim point.
a. What was the bullet's flight time?
b. What was the bullet's speed as it left the barrel?

My second problem: A projectile is fired with an initial speed of 30 m/s at an angle of 60(degrees) above the horizontal. The object hits the ground 7.5 s later.
a. How much higher or lower is the launch point relative to the point where the projectile hits the ground?
b. To what maximum height above the launch point does the projectile rise?
c. What are the magnitude and direction of the projectile's velocity at the instant it hits the ground?

The final problem is: A stunt man drives a car at a speed of 20 m/s off a 30-m-high cliff.
The road leading to the cliff is inclined upward at an angle of 20(degrees).
a. How far from the base of the cliff does the car land?
b. What is the car's impact speed?(I didn't understand what is an impact speed)

2. Relevant equations
The range equation: d=((Vi)2.sin(2@))/g
Xf= Xi + Vixt
yf= yi + Viyt - 1/2g(t)2
g= 9.8 m/s2

3. The attempt at a solution
I didn't understand these three problems, even i didn't know how to draw them, so plz help me because i'm tired from the other problems that i had solve.

2. Sep 28, 2007

AvrGang

I'm logged on for any discussion

3. Sep 28, 2007

NeoDevin

First Question: What was the initial vertical component of the bullet's velocity, what was the vertical distance traveled, and what was the vertical accelleration? Do you have an equation which relates these three things to find time?

Second Question: Again, you know the all the information about the vertical components of the motion, except the final position, can you solve for this?

Third Question: Impact speed is the speed it has just before hitting bottom. This question follows similar ideas to the last two.

4. Sep 28, 2007

chaoseverlasting

There are two things to consider here: the bullets' horizontal velocity and its vertical velocity. There is no acceleration on the bullet after it has left the rifle's barrel. Therefore, the only acceleration is in the downward direction due to the force of gravity given by g=9.8m/s^2.

Write out the equations of motion for the x and y axes respectively and solve them for the time taken.

5. Sep 28, 2007

AvrGang

Aha..... Thanks for helping me , i'll try now

6. Sep 28, 2007

FedEx

Hi,
The first question is easier than the other two.For the first one you will have to consider that the distance covered in the y direction is 2cm.And the distance travelled in the x direction is 50 m.Now i think that you must do the question by applying the normal laws of kinematics and projectile motion.

For the second question. Find out the time taken by the projectile to travel the whole distance. If the time is greater than 7.5 seconds than it means that after 7.5 seconds the projectile is still in motion and it is above the launch point by a certain amount of height.That you can solve by the equation

$$d_y = v_0\sin\theta{t} + \frac{at^2}{2}$$

Now for the third question there is an equation for it. Do you know that equation it is different than the three conventional equations which we have.

7. Sep 28, 2007

AvrGang

I wish i know it

but thank you for helping me

8. Sep 28, 2007

FedEx

Dont worry.

So did you get the first two. If yes than we may discuss the equation for the third question

9. Sep 28, 2007

AvrGang

ok i solved the first one

I'm now solving the second one

10. Sep 28, 2007

AvrGang

d_y = v_0\sin\theta{t} + \frac{at^2}{2}
d_y = 30sin60(7.5)-0.5(9.8)(7.5)^2
d_y = -80.8m

How i can know if it's higher or lower?

11. Sep 28, 2007

AvrGang

I solved question a) in the second question

12. Sep 28, 2007

AvrGang

plz someone help me in b)To what maximum height above the launch point does the projectile rise? In the second problem

13. Sep 28, 2007

FedEx

I think that there is something wrong.Hold on.

14. Sep 28, 2007

AvrGang

Thank god I finally solved the first two questions, thank you guys for helping me

Ill work now on the third question which is:

A stunt man drives a car at a speed of 20 m/s off a 30-m-high cliff.
The road leading to the cliff is inclined upward at an angle of 20(degrees).
a. How far from the base of the cliff does the car land?
b. What is the car's impact speed?(I didn't understand what is an impact speed)

I think this question is harder that the first two
I'll work now on it
but plz if you have any hint plz tell me
I'll still outline

15. Sep 28, 2007

AvrGang

the second question is higher because of the (-) sign which means in the opposite direction of our convention( i choose the up direction to be positive)

16. Sep 28, 2007

FedEx

It is lower. Because if you count the time where the whole land is of same level than you will find that t=5.29 seconds.Hence the ball might have touched the ground after 5.29 seconds if the land was levelled . But it has taken some more time and hence it is lower than the shooting point.

Last edited: Sep 28, 2007
17. Sep 28, 2007

FedEx

So have you got the correct answers of part 1 and part 2.

The part three is not what i thought. It does not need any complicated equation. But now looking at the problem i feel that either the question is incomplte or i think that i have misunderstood the question.

Is the take of velocity of the car 20m/s? Or is it the velocity when the car is during the flight and that also when it is exactly above the highest point of cliff?

18. Sep 28, 2007

AvrGang

I finished my assignment
Thank you all