- #1

silvaramaxwel

- 2

- 0

## Homework Statement

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

## Homework Equations

1. x=v

_{0x}* t

2. y = v

_{0y}* t - 1/2gt

^{2}

3. v

_{y}= v

_{0y}- gt

4. v

_{y}

^{2}= v

_{0y}

^{2}- 2gΔy

## The Attempt at a Solution

I assumed at the top of the leap, v

_{y}=0 m/s, so I used equation 4 to find the initial y velocity (0m/s = v

^{2}- 2(9.8m/s

^{2})(2.1m) which gave me v

_{y}= 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.