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Projectile Motion, finding minimum speed

  • #1

Homework Statement


A gazelle leaps over a 2.1m fence. Assuming a 45° takeoff angle, what is the minimum speed?


Homework Equations


1. x=v0x * t
2. y = v0y * t - 1/2gt2
3. vy = v0y - gt
4. vy2 = v0y2 - 2gΔy


The Attempt at a Solution


I assumed at the top of the leap, vy=0 m/s, so I used equation 4 to find the initial y velocity (0m/s = v2 - 2(9.8m/s2)(2.1m) which gave me vy = 6.416 m/s)

I put the final y velocity into equation 3 to find t=0.6547 s.

Then I used 2.1m/tan 45° to find the x distance (2.1 m) and used that distance plus the time I found and plugged them into equation 1 to get the initial x velocity of 3.208 m/s.

Finally, I used the Pythagorean equation to find the actual initial velocity... The actual answer is 9.073 m/s, but I can't seem to work the problem out to that. I have a lot of problems with finding the initial minimum speed/velocity when given the final distances (even though I can find the range and maximum height when given the initial velocity) so any help would be greatly appreciated.
 

Answers and Replies

  • #2
1,065
10
assumed at the top of the leap, vy=0 m/s, so I used equation 4 to find the initial y velocity (0m/s = v2 - 2(9.8m/s2)(2.1m) which gave me vy = 6.416 m/s)
--------------------------
You got the vertical component.
The velocity is the sum of 2 components of x and y.
Use Pythagoras' Theorem to solve for v.
 
  • #3
Right I got that it was the y velocity, and I tried to find the x component and do the Pythagorean equation... but I'm seeing now that the easiest thing to do is to use the y velocity/sin 45 to get the right answer.

Thanks.
 

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