Physics Projectile Motion Help (1 Viewer)

[luo_ge]

 

[strawberrye]

In the future, please post this in the appropriate sub-forum

 

[Green Yoda]

I DELETED MY OTHER POST CUZ I DID IT BY INSPECTION AND WAS WRONG AF LOL

13)
Δy=50
u(x)=1060cos(0)=1060m/s
u(y)=0

Δy=u(y)+0.5a(y)*t^2
Δy=0+4.9t^2
t=3.19s

Δx=u(x)*t
Δx=1060*3.19
Δx=3386.01m

 

[Green Yoda]

14)
Lets Divide this trajectory in T1 and T2 where T1= time from start to max height and T2 is from max height to rest.

u(x)=1060cos(45)=749.53m/s
u(y)=749.53m/s


T1:
v(y)=u(y)+a(y)*t
0=749.53-9.8*t
t=749.53/9.8
t=76.48s

Δy=(u(y)*t+0.5*a(y)*t^2)+50
Δy=(749.53*76.48-4.9*(76.48)^2)+50
Δy=28663.02+50
Δy=28713.02m

T2:
Δy=0.5*a(y)*t^2
28713.02=4.9*t^2
t=75.55s

----------
T.O.F=153.03s
Δx=u(x)*t
Δx=114700.58m <---- THIS SEEMS COMPLETELY OFF, LEMME KNOW IF I'VE MADE A MISTAKE

 

[pikachu975]

14)
Lets Divide this trajectory in T1 and T2 where T1= time from start to max height and T2 is from max height to rest.

u(x)=1060cos(45)=749.53m/s
u(y)=749.53m/s


T1:
v(y)=u(y)+a(y)*t
0=749.53-9.8*t
t=749.53/9.8
t=76.48s

Δy=(u(y)*t+0.5*a(y)*t^2)+50
Δy=(749.53*76.48-4.9*(76.48)^2)+50
Δy=28663.02+50
Δy=28713.02m

T2:
Δy=0.5*a(y)*t^2
28713.02=4.9*t^2
t=75.55s

----------
T.O.F=153.03s
Δx=u(x)*t
Δx=114700.58m <---- THIS SEEMS COMPLETELY OFF, LEMME KNOW IF I'VE MADE A MISTAKE

Testing with the u^2 sin2theta / g formula it gets the same answer as you so I think it's right.

 

[Green Yoda]

Testing with the u^2 sin2theta / g formula it gets the same answer as you so I think it's right.

Yeah couldn't see any mistake..but that number looks unreasonable

 

[pikachu975]

Yeah couldn't see any mistake..but that number looks unreasonable

But if you think about it... 749.5 metres per second horizontally is heaps and since it's 45 degrees it'll have a sizeable vertical displacement too, making the TOF larger.