Free Fall and a Projectile motion question

1. Dec 7, 2016

rockcandy

1. The problem statement, all variables and given/known data
1. A diver jumps from a height of 10m leaving with a velocity of 8m/s find the time for which the diver is in the air and the speed at which the diver enters the water?

2. A ball is launched from point A on a horizontal plane with an angle of 55*. It passes two walls both 2m high. Wall one is 2m from point A. Find the speed of the ball and the Wall two distance from point A

The attempt at a solution

1. Vy= 8m/s dy = 10m ay=-9.8m/s2

Vfy2 = Viy2 + 2ayd

Vy=√(8m/s)2 + 2(-9.8m/s2)(-10m)
Vy = 16.1 m/s

-10 m = 8 m/s t + 1/2 (-9.8m/s2) t2
0 = -4.9 m/s2 t2 + 8m/s t + 10m

t = 2.45 s or - 0.82 s

2. Vx = VicosΘ and Vy = VisinΘ h =2m and horizontal distance = 2m

dy= Viyt + 1/2 at2 dx = Vixt + 1/2at2
since horizontal acceleration = 0 ; dx = Vixt
dy= Viyt + 1/2 at2 dx = Vixt ; 2 m = VicosΘ t
2 = VisinΘ t - 1/2 (9.8m/s2) t 2
2 = VisinΘ (2m /VicosΘ) - 1/2 (9.8m/s2)(2m /VicosΘ)2
2 = 2m tan Θ - 4.9m/s2 (12.15 m 2 /Vi2 )
2 = 2.85 m - 59.535 / Vi2
Vi = √70.04
Vi = 8.36 m/s

Im stuck on how to get the distance assuming this is correct. If I use the max range formula

x = Vi2 sin2Θ / g
x = (8.36m/s)2 sin 2(55) / 9.8 m/s2
x = 6.69 m

If I take half this distance ( 6.69m)(.5) = 3.34 m
I know that it is reaching a height of 2 m twice, and given that the parabola is symmetrical I know that on the opposite side it is also 2m away from Wall 2.
3.34 m - 2 m = 1.34 m , which the remaining distance between the wall to the half the max range.

therefore 1.34m x 2 = 2.68m which is the total distance between Wall 1 and Wall 2

So to find the distance from Wall 2 to Point A just add the total = 2.68m + 2 m = 4.68m

Last edited: Dec 7, 2016
2. Dec 7, 2016

PeroK

Does $21.6s$ not seem a long time to be in the air?

3. Dec 7, 2016

rockcandy

Youre right... my math was off
its t = 2.45s or -0.82

assuming everything else is correct

4. Dec 7, 2016

PeroK

You have to make your mind up about whether it's $2.45s$ or $-0.82s$. Do you understand why there are two solutions?

In question 2, are you supposed to assume that the ball just clears each wall?

5. Dec 7, 2016

rockcandy

The time is t = 2.45 s
For #2 yes you assume it clears each wall

6. Dec 7, 2016

PeroK

I really like your approach to question 2. Using the range (or you could have used the distance to the highest point) and then the symmetry of the parabola was very neat.

One suggestion is to try to use algebra a little more and not be so quick to plug in the numbers.

7. Dec 7, 2016

rockcandy

Thanks ; But did I solve the 2 problems correctly?

8. Dec 7, 2016

PeroK

Yes, of course!

9. Dec 7, 2016

PeroK

PS I think the simpler approach was to find the two times that $y = 2m$ and the use the x-velocity to get the distance to the second wall.

10. Dec 7, 2016

rockcandy

Thanks for the help , I drew a diagram and I see what you mean.

11. Dec 7, 2016

PeroK

Here's a thing. I reckon that if $h$ is the height of the walls, $a$ is the distance to the first wall and $b$ the distance to the second wall, then:

$b = \frac{ah}{a\tan \theta - h}$

In this case $a = h = 2m$ so:

$b = \frac{2}{\tan \theta - 1}m$

Is that interesting?