An athlete jumped a pole vault that was over 2 m high . Her horizontal distance was 5m. What was her initial speed?
5 years ago
Answered By James S
There are two main formulas for projectile motion:
1) vf2 = vi2 + 2ad
2) d = vit + 1/2*at2
We are given two distances, a maximum height of 2m (let's call it 'h') and the horizontal distance from the initial launch to that point (let's call it 'b'). We can draw it like below.
We need to find our vi, however we need to first break it down into our x and y components.
In y (vertical):
a=-9.81m/s2. It's negative since it's acting opposite of our movement, ie. it's slowing down our projectile.
At our maximum height, our vf = 0m/s. So, plugging in to eq. 1):
vfy2 = 0 = viy2 + 2(-9.81m/s2)*(2m)
So, viy = 6.27m/s
We can use Eq 2 to solve for 't' now.
d=h=2m=(6.27m/s)2*t + 1/2*(-9.81m/s2)*t2
Rearrange in form of quadratic equation:
4.905t2-6.27t+2=0
Solve for t using quadratic formula:
t = $\frac{\left(-\left(-6.27\right)\pm\sqrt{\left(\left(-6.27\right)^2-4\left(4.905\right)\left(2\right)\right)}\right)}{2\left(4.905\right)}$(−(−6.27)±√((−6.27)2−4(4.905)(2)))2(4.905) = 0.64
Now we can plug into Eq 2 and use the x-coordinates to determine vix.
Remember, in the x-direction, a=0 and d=5
5m=vix(0.64s) + 0
vix= 7.81m/s
Total vi = (vix2+viy2)1/2 = ((7.81m/s)2+(6.27m/s)2)1/2 = 10.02m/s
5 years ago
Answered By James S
There are two main formulas for projectile motion:
1) vf2 = vi2 + 2ad
2) d = vit + 1/2*at2
We are given two distances, a maximum height of 2m (let's call it 'h') and the horizontal distance from the initial launch to that point (let's call it 'b'). We can draw it like below.
We need to find our vi, however we need to first break it down into our x and y components.
In y (vertical):
a=-9.81m/s2. It's negative since it's acting opposite of our movement, ie. it's slowing down our projectile.
At our maximum height, our vf = 0m/s. So, plugging in to eq. 1):
vfy2 = 0 = viy2 + 2(-9.81m/s2)*(2m)
So, viy = 6.27m/s
We can use Eq 2 to solve for 't' now.
d=h=2m=(6.27m/s)2*t + 1/2*(-9.81m/s2)*t2
Rearrange in form of quadratic equation:
4.905t2-6.27t+2=0
Solve for t using quadratic formula:
t = $\frac{\left(-\left(-6.27\right)\pm\sqrt{\left(\left(-6.27\right)^2-4\left(4.905\right)\left(2\right)\right)}\right)}{2\left(4.905\right)}$(−(−6.27)±√((−6.27)2−4(4.905)(2)))2(4.905) = 0.64
Now we can plug into Eq 2 and use the x-coordinates to determine vix.
Remember, in the x-direction, a=0 and d=5
5m=vix(0.64s) + 0
vix= 7.81m/s
Total vi = (vix2+viy2)1/2 = ((7.81m/s)2+(6.27m/s)2)1/2 = 10.02m/s
:)
5 years ago
Answered By Sosimo H
Attached Whiteboard:
Play Drawing5 years ago
Answered By Sosimo H
Let's first analize vertical motion
v0=0
a=9.8 m/s2
y0 = 0
y=2 m
y0=0 when it is vertical
Y= Y0 +V0t + 1/2 at2
t2=2y/a t= $\sqrt{\frac{2.2}{9.8}}=\sqrt{\frac{4}{9.8}}=.64sec$√2.29.8 =√49.8 =.64sec
Horizontal
x= x?+v0 t+1/2at2
x0=0
a=0
x=v0t , v0=x/t = 5 m/ 0.64 sec =7.81 m/s