Calculating Projectile Motion for a Catapult with Limited Data

[thedarklite]

Hi Guys
hope u r feeling good
i just finished my catapult project
and the problem is that i have only two given
and the problem is that i have only two given values the $$\Delta T = 1.1 sec.\Delta_{d}x=2.2 m$$, and the angle is 70' degrees. and nothing else.

http://tinypic.com/r/2ylqpux/7

i have tried to use the projectile motion equations but there was so many unknowns
what is confusing me more that i have the catapult powered by a compression spring and i don't know if that is effecting the calculations or not.

pleasezz i need help at this because I'm totally lost

 

[gneill]

Hi thedarklite, welcome to Physics Forums.

Please use the thread template when you post a question.

You haven't stated the complete problem; what is it you're trying to find/solve?

 

[thedarklite]

sorry for that gneill
but all what i need to find is the deltaYd and the Vyf and initial too but the problem this is my first time doing the catapult thingy and the problem is that i have a weak understanding of physics that is why i don't know how to approach this problem

 

[gneill]

Can you write the two equations of motion that describe the x and y motions of the projectile? The important thing to remember is that the x and y components (that is, the horizontal and vertical motions) of the projectile are independent of each other.

 

[thedarklite]

i think i am suppose to find the final and initial velocity of the y component i think i also need to find the maximum height of the y component
and i don't know which equations to use

 

[gneill]

What equations of motion do you know? For a constant velocity v, what is the expression for the distance versus time? How about when there's an initial velocity and an acceleration?

 

[thedarklite]

so that is the solution that i tried

Variables:
∆dx=2.1 m ∆t=1.1 s V_ix= ? V_iy= ? V_fy= ? a_y= ?
∆dy=0.0275 m
V_ix= ∆dx⁄∆t V_ix= (2.1 m)⁄(1.1 s)
∴V_ix= 1.9 m⁄s
V_iy= V_ix tan⁡〖70°〗 V_iy=(1.9 m⁄s) tan⁡〖70°〗,
∴V_iy=5.2 m⁄s
∆dy= V_iy ∆t+ 1/2 a_y (∆t)^2 0.0275m=(5.2 m⁄s)(1.1s)+ 1/2 a_y (1.1s)^2
∴a_y=-9.4 m⁄s^2
〖V_fy〗^2= 〖V_iy〗^2+2a_y ∆dy V_fy= √((5.2 m⁄s)^2+2(-9.4 m⁄s^2 )(0.0275m) )
∴V_fy=5.1 m⁄s since V_ix= V_fx ∴V_fx=1.9 m⁄s
V= √(〖V_fy〗^2+ 〖V_fx〗^2 ) V= √((5.1)^2+ (1.9)^2 ) ∴V=5.4 m⁄s
tan^(-1)⁡〖5.1/1.9〗= 70°,tan^(-1)⁡〖1.9/5.1〗=20°
∴V=5.4 m⁄s [W 70° N] or V=5.4 m⁄s [S 20° E]

 

[gneill]

Your original problem statement seemed to indicate that the total x-distance was 2.2m, not 2.1m. So the velocity in the x-direction should be 2.2m / 1.1s = 2.0 m/s.

With the x-component of the velocity and the angle, you should be able to find the initial y-component of the velocity.