# Projectile Motion, finding minimum speed

## 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.

Related Introductory Physics Homework Help News on Phys.org
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)
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You got the vertical component.
The velocity is the sum of 2 components of x and y.
Use Pythagoras' Theorem to solve for v.

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.