Projectile Motion at an Angle: Solving for Time, Distance, and Final Velocity

In summary, the projectile hits the ground 3.1 seconds after being shot from the cliff. It has a final speed of 72.47 m/s.
  • #1
Precepts
14
0
Hi, I need someone to check if I did this right. As I got the q's off a PDF with no answers.

Homework Statement



A projectile is shot from the edge of a cliff 125 m above the ground. It has an initial velocity 50.0 m/s 37.0° above the horizontal at the launch point. a) Determine the time taken to reach the ground. b) How far is this point from the base of the cliff? c) What is the magnitude and direction of the final velocity just before the point of impact?

It doesn't specify what value to use for gravity, I assumed g, or a, is = -9.8m/s^2
(Doesn't matter since there's no answers anyway)

Homework Equations



v = u + at
s = ut (horizontal sense)

The Attempt at a Solution

I used this to find the time for a)

Vv = Uv + Avt
0 = 50 * sin37 - 9.8t
9.8t = 50 * sin37
t = 3.1s

b) Sh = Uh * t
= 50 * cos37 * 3.1
Sh = 123.79m

c) Vertical component before impact:

Vv = Uv + Avt
= 60.47m/s

I then used Pythagoras' to find v = 72.47m/s
and theta = 56'33'' above the horizontal. I feel that I went wrong somewhere, can someone check if these are right/close. Not sure If I copied everything down correctly either, my page is a bit of a mess.edit - I just noticed I didn't once use the vertical distance, I think I'll have another look at it.

edit 2 - Sv = Uvt + 1/2At^2
125 = 50 sin 37t -4.9t^2
= 30.1t - 4.9t^2

-4.9t^2 + 30.1t - 125 = 0
4.9t^2 - 30.1t +125 = 0

And... I'm getting math error when using quadratic formula. Not quite sure where to go from here...I realize now that the value for t that I had would only reach to the same height on the other side of the parabola... I'll have a look at this question again later.
 
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  • #2
a)Vv = Uv + Avt
0 = 50 * sin37 - 9.8t
9.8t = 50 * sin37
t = 3.1s

The vertical component of the speed is not 0 when it hits the ground!
 
  • #3
Basic_Physics said:
a)Vv = Uv + Avt
0 = 50 * sin37 - 9.8t
9.8t = 50 * sin37
t = 3.1s

The vertical component of the speed is not 0 when it hits the ground!


Right, lol.

Can you find t? I tried to incorporate the height and use the quadratic formula, but it wouldn't solve. Kind of stuck here.
 
  • #4
Found it. I had to use a negative displacement for the height... I was getting a negative square root prior. Wow...
 
  • #5


Hi there,

I am a scientist and I would be happy to help you with your problem. Based on the information provided, it seems like you have a good understanding of the concepts of projectile motion and have attempted to solve the problem correctly. However, there are a few mistakes in your calculations that I would like to point out.

Firstly, in part a), you have correctly used the equation Vv = Uv + Avt to find the time. However, you have made a mistake in your calculation of Vv. It should be Vv = 50 sin(37) - 9.8t, not 50 * sin37 - 9.8t. This will give you a slightly different value for time, which will affect your calculations in the later parts.

Secondly, in part b), you have used the equation Sh = Uh * t, which is incorrect. This equation only applies for motion in the horizontal direction. For motion in the vertical direction, you need to use the equation Sv = Uv * t + 1/2 * Av * t^2. This will give you a different value for the horizontal distance.

Finally, in part c), you have correctly calculated the vertical component of the final velocity, but your calculation for the magnitude and direction is incorrect. To find the magnitude, you need to use the equation v = √(Vv^2 + Vh^2). Also, the direction of the final velocity should be 56°33' above the horizontal, not 56'33''.

I hope this helps you in correcting your calculations. Keep up the good work and don't hesitate to ask for clarification if needed. Good luck!
 

Related to Projectile Motion at an Angle: Solving for Time, Distance, and Final Velocity

1. What is projectile motion at an angle?

Projectile motion at an angle refers to the motion of an object that is launched into the air at an angle, rather than directly upwards or horizontally.

2. How is the trajectory of a projectile at an angle different from that of a horizontally launched projectile?

The trajectory of a projectile at an angle is a parabolic curve, while the trajectory of a horizontally launched projectile is a straight line.

3. What factors determine the range of a projectile at an angle?

The range of a projectile at an angle is determined by the initial velocity, launch angle, and the acceleration due to gravity.

4. Can the maximum height of a projectile at an angle be calculated?

Yes, the maximum height of a projectile at an angle can be calculated using the equation h = (v*sinθ)^2 / 2g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

5. How does air resistance affect projectile motion at an angle?

Air resistance can affect the trajectory and range of a projectile at an angle by slowing it down and causing it to fall shorter than expected. It also causes the projectile to deviate from a perfect parabolic path.

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