Initial Velocity of Projectile Motion, given displacement and initial angle?

[allure]

Homework Statement

A grasshopper jumps 1.00 meters from rest, with an initial velocity at a 45.0° angle with respect to the horizontal. Find (a) the initial speed of the grasshopper and (b) the maximum height reached.

Homework Equations

v_x = v_0xt = v₀cos θ₀

v_y = v₀ sin θ₀ - gt

The Attempt at a Solution

I was unsure what to do, so I resolved for the components, resulting in

v_0x = v₀ cos 45°

v_0y = v₀ sin 45°

I apologize for my cluelessness - this my first semester in physics (at all), and I am having trouble finding where to go about this problem. Thank you in advance.

 

[SHISHKABOB]

is that 1.00 supposed to be in meters? Is the grasshopper jumping 1.00 meter forward across the ground?

also, when the grasshopper reaches the peak of its arc through the air, what is the magnitude of the *vertical component* of the velocity? (this is a question that is trying to push you in the right path, the first one is clearing up some confusion on what the problem is asking)

 

[allure]

Yes, meters - sorry. I have edited the above.

 

[allure]

is that 1.00 supposed to be in meters? Is the grasshopper jumping 1.00 meter forward across the ground?

also, when the grasshopper reaches the peak of its arc through the air, what is the magnitude of the *vertical component* of the velocity? (this is a question that is trying to push you in the right path, the first one is clearing up some confusion on what the problem is asking)

The velocity of the peak of the arc would be 0. I tried to look for a way to use this to find the initial speed but I'm stuck, is there a way I can use this information to do that?

 

[SHISHKABOB]

well so you can solve for the t it takes to get to that point

and so then if that's the amount of time it takes to get to the *peak*, how long does it take the grasshopper to go the full distance?

 

[allure]

So far, I've solved using V_y sin θ₀ - gt
to get

^tmax = v₀sin 45°/9.8

t_total = 2(v₀ sin 45°/9.8)

Without the initial velocity, I'm not sure how (if I can) to simplify any further, or solve for it..

 

[aftershock]

45 degrees is right in the middle of completely horizontal and completely vertical. You might feel like the horizontal and vertical components should be the same, and indeed they are. cos45=sin45 Might as well take V_ox and call it V_oy

X = V_oyt ,where X is the horizontal distance traveled.

V_yf = _Voy -gt

However V_yf is right before hitting the ground... it must have equal magnitude but opposite direction of <sub>Voy so:

-Vyo = Voy -gt

So you have 2 equations, two unknowns and can solve.</sub>

 

[SHISHKABOB]

so if you know that the time taken for the grasshopper to end its jump is t_total = 2(v₀ sin 45°/9.8)

and that it went 1 meter

and you know that

x = v_0,xt = v₀cos(θ)t

what could you do with that

 

[mishek]

Hello, check Range of trajectory.