Projectile motion, ideal angle

[Taylan]

Homework Statement

A frog jumps at t=0s and follows a projectile motion. The maximum height he reaches is 0.45m. The air resistance can be neglected.

a) What is the initial speed of the frog in y-direction and how long is the total time until he lands on the ground?

b) At which degree should he jump from the horizontal plane, to be able to reach a horizontal range of 3m?

Homework Equations

v^2=u^2+2as
v=d/t[/B]

The Attempt at a Solution

In part a I wrote down what I know about the vertical motion.
=> finding initial speed in y-direction
s=0.45m
u=?
v=0m/s
a= -9.81m/s^2

v^2=u^2+2as
u=2.97m/s

=> finding the total time of the flight
s=0m
u=2.97m/s
a=-9.81m/s^
T=?
s=ut+1/2at^2
t=0.61sFor part b I am confused. So the velocity in horizontal direction is always the same so I can use v=d/t
I know d=3m. I don't know the initial velocity or time it would take for a range of 3m. Can you give me some tips?

 

[PeroK]

b) At which degree should he jump from the horizontal plane, to be able to reach a horizontal range of 3m?For part b I am confused. So the velocity in horizontal direction is always the same so I can use v=d/t
I know d=3m. I don't know the initial velocity or time it would take for a range of 3m. Can you give me some tips?

For part b) you need to introduce the angle, $\theta$ say, as a variable in your equations.

 

[Taylan]

For part b) you need to introduce the angle, $\theta$ say, as a variable in your equations.

in horizontal plane v=d/t v is always the same which is initial speed,u, times cosθ.

ucosθ= d/t

d= ucosθ * t

now I only know d (3m) but don't know u and t. t found in part a was for that particular case where max height was 0.45m so it seems ı can't use it.
can you give me more tips please?

 

[PeroK]

in horizontal plane v=d/t v is always the same which is initial speed,u, times cosθ.

ucosθ= d/t

d= ucosθ * t

now I only know d (3m) but don't know u and t. t found in part a was for that particular case where max height was 0.45m so it seems ı can't use it.
can you give me more tips please?

Can you calculate $t$ somehow? In general, in a problem like this you want to eliminate $t$ from your equations.

 

[Taylan]

Can you calculate $t$ somehow? In general, in a problem like this you want to eliminate $t$ from your equations.

it seemed to me I can't. I only know the horizontal range in b.

 

[PeroK]

it seemed to me I can't. I only know the horizontal range in b.

First, every angle will result in a different range for the same initial velocity. If you fire it straight up, it lands back on the same point. If you fire it at an angle it goes some distance $R$. $R$ is maximised when the angle is $45°$. After that $R$ reduces again until the angle is zero and it effectively never gets off the ground.

So, there must be at most two solutions for a range of $R = 3m$ for a given initial velocity. Therefore, you have enough information to solve the problem.

Now, what determines the time of flight?

 

[Taylan]

First, every angle will result in a different range for the same initial velocity. If you fire it straight up, it lands back on the same point. If you fire it at an angle it goes some distance $R$. $R$ is maximised when the angle is $45°$. After that $R$ reduces again until the angle is zero and it effectively never gets off the ground.

So, there must be at most two solutions for a range of $R = 3m$ for a given initial velocity. Therefore, you have enough information to solve the problem.

Now, what determines the time of flight?

ok thanks! the initial velocity determines the time of flight.

 

[Taylan]

this is what I did so far:

--vertical--
s=0
u= usinθ
v= -usinθ
a= -9.81m/s^

s= ut+1/2at^2
t= (usinθ)/4.905....(1)

--horizontal--
d=3m
v= ucosθ
t=d/v= 3/ucosθ ....(2)

combining (1) and (2);

u/2 * sin(2θ)=3(4.905)

but now I'm unable to find θ since I don't know the initial velocity in z direction, u .

 

[PeroK]

this is what I did so far:

--vertical--
s=0
u= usinθ
v= -usinθ
a= -9.81m/s^

s= ut+1/2at^2
t= (usinθ)/4.905....(1)

--horizontal--
d=3m
v= ucosθ
t=d/v= 3/ucosθ ....(2)

combining (1) and (2);

u/2 * sin(2θ)=3(4.905)

but now I'm unable to find θ since I don't know the initial velocity in z direction, u .

You calculated $u$ in part a). That should give you $\theta$.

 

[Taylan]

You calculated $u$ in part a). That should give you $\theta$.

but that u ( speed in vertical direction) was for the particular case where my max height is 0.45m right? Now with a range of 3m, the speed in vertical direction will be different I thought

 

[PeroK]

but that u ( speed in vertical direction) was for the particular case where my max height is 0.45m right? Now with a range of 3m, the speed in vertical direction will be different I thought

$u$ is the initial speed, which you calculated in part a). You are to assume the frog jumps with the same initial speedeach time.

$u \sin \theta$ is the vertical component.

 

[Taylan]

$u$ is the initial speed, which you calculated in part a). You are to assume the frog jumps with the same initial speedeach time.

$u \sin \theta$ is the vertical component.

in part a, speed of frog in vertical direction was asked and ı named it u. actually ı should have named it usinθ to be more clear. so :
usinθ=2.97m/s from part a. I still don't know the u. right?

 

[PeroK]

in part a, speed of frog in vertical direction was asked and ı named it u. actually ı should have named it usinθ to be more clear. so :
usinθ=2.97m/s from part a. I still don't know the u. right?

Hmm. I'm not sure now what the questioner intended.

1) What I was assuming was that the first jump was a vertical jump. Then, the second jump was a jump at an angle with the same take-off speed.

2) But, it could be the same vertical speed in each case.

It's not clear from the question, but it's more logical to assume 1).

 

[Taylan]

It is stated in the question that both jumps are angular. So two projectiles motions

 

[PeroK]

It is stated in the question that both jumps are angular. So two projectiles motions

It's not stated in the question you posted.

Anyway, even a frog can't high-jump and long-jump at the same time!

If you want to assume the same vertical speed, then the question is quite simple, as the time of flight is the same in both cases.

 

[Taylan]

It's not stated in the question you posted.

Anyway, even a frog can't high-jump and long-jump at the same time!

If you want to assume the same vertical speed, then the question is quite simple, as the time of flight is the same in both cases.

Yes it is not stated in the question I posted, I wasn't clear there. I mean it was stated in the original question. And can we get any solutions without that assumption somehow?

 

[PeroK]

Yes it is not stated in the question I posted, I wasn't clear there. I mean it was stated in the original question. And can we get any solutions without that assumption somehow?

You have to decide on one assumption or other. You could do both versions, of course.

 

[haruspex]

Yes it is not stated in the question I posted, I wasn't clear there. I mean it was stated in the original question. And can we get any solutions without that assumption somehow?

My feeling is that the second part of the question is just badly worded, that it should read "if he reached 3m horizontally, what angle did he jump at?"

 

[Taylan]

My feeling is that the second part of the question is just badly worded, that it should read "if he reached 3m horizontally, what angle did he jump at?"

Yes , the original question was in german and i tried my best to translate it

 

[PeroK]

Yes , the original question was in german and i tried my best to translate it

Actually, I just noticed that if the frog can only jump $0.45m$ vertically, then it's impossible to jump $3m$ horizontally. So, the question must assume a constant vertical take-off speed and a variable horizontal speed.

Or, of course, parts a) and b) are talking about the same jump. That makes more sense.