What is the Ideal Angle for a Frog's Jump to Reach 3 Meters Horizontally?

• Taylan
In summary: ...the initial speed in y-direction. s=0.45mu=2.97m/sv=0m/sa=-9.81m/s^2
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?

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?

Taylan said:
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.

PeroK said:
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?

Taylan said:
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.

PeroK said:
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.

Taylan said:
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?

PeroK said:
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.

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 .

Taylan said:
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##.

PeroK said:
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

Taylan said:
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.

PeroK said:
##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?

Taylan said:
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).

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

Taylan said:
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
PeroK said:
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?

Taylan said:
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.

Taylan said:
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?"

haruspex said:
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

Taylan said:
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.

1. What is projectile motion?

Projectile motion is the movement of an object through the air or space under the influence of gravity. It follows a curved path, also known as a parabolic path, due to the combination of its horizontal and vertical velocities.

2. What is the ideal angle for a projectile?

The ideal angle for a projectile is 45 degrees. This angle maximizes the horizontal distance traveled by the projectile, known as the range, for a given initial velocity and height.

3. How is the ideal angle for a projectile determined?

The ideal angle for a projectile is determined using the formula 𝟐𝒕𝒉𝒆𝒕𝒂𝒏𝒈𝒆𝒏𝒈𝒍𝒆 = tan(𝜃), where 𝜃 represents the angle and 𝒕𝒉𝒆𝒕𝒂𝒏𝒈𝒆𝒏𝒈𝒍𝒆 represents the horizontal distance traveled. By taking the derivative of this formula and setting it equal to zero, we can find the angle that maximizes the horizontal distance, which is 45 degrees.

4. Can the ideal angle for a projectile change?

Yes, the ideal angle for a projectile can change depending on factors such as the initial velocity and height, as well as external factors such as air resistance and wind. However, in a vacuum with no external factors, the ideal angle will always be 45 degrees.

5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the projectile, causing it to have a shorter range and a different ideal angle. It can also cause the projectile to follow a non-parabolic path, making it more difficult to predict its trajectory and landing point.

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