Projectile Motion Question (1 Viewer)

[luo_ge]

Can someone help me find the final velocity, max height, range and time of flight when given:

vertical displacement: 0.2m
horizontal velocity: 1.49m/s
vertical velocity: 1.99m/s

Thanks.

 

[Green Yoda]

What type of a projectile motion is it? Symmetrical?

 

[Green Yoda]

Need more info on the actual motion of the projectile...where its thrown from and in what direction

 

[luo_ge]

What type of a projectile motion is it? Symmetrical?

it's not symmetrical. it's thrown in a parabola from left to right. it starts off 0.2m (20cm) from the ground and ends at 0m above the ground

 

[Green Yoda]

it's not symmetrical. it's thrown in a parabola from left to right. it starts off 0.2m (20cm) from the ground and ends at 0m above the ground

Alright, but make sure you specify what the vertical displacement is..projectile or the starting point.

 

[Green Yoda]

Can someone help me find the final velocity, max height, range and time of flight when given:

vertical displacement: 0.2m
horizontal velocity: 1.49m/s
vertical velocity: 1.99m/s

Thanks.

Are these Initial or Final?? Well for horizontal component it doesn't matter but for vertical component it does.

 

[Green Yoda]

ASSUMING YOU HAVE GIVEN DATA FOR INITIAL VELOCITIES:
Separate this motion into T1 and T2 where T1 is from start to max height and T2 is from max height to stop.

T1:
v(y)=u(y)+a(y)t
0=1.99-9.8(t)
t(max)=0.2s

Δy=(u(y)t+0.5a(y)t^2)+0.2
Δy=(1.99*0.2-4.9*0.2^2)+0.2
Δy=0.2+0.2=0.4m

T2:
Δy=u(y)t+0.5a(y)t^2
0.4=0+4.9*t^2
t=0.29

THEREFORE TIME OF FLIGHT = 0.2+0.29=0.49s

Δx=u(x)*t
Δx=1.49*0.49
Δx=0.73m

v(y)=u(y)+a(y)*t (whole trajectory)
v(y)=1.99+9.8*0.49
v(y)=6.79m/s

*Using Pythagoras and trig which are assumed and elementary so I wont bother putting in the working*
Final Velocity= 6.95m/s 78° below the horizontal

 

[luo_ge]

ASSUMING YOU HAVE GIVEN DATA FOR INITIAL VELOCITIES:
Separate this motion into T1 and T2 where T1 is from start to max height and T2 is from max height to stop.

T1:
v(y)=u(y)+a(y)t
0=1.99-9.8(t)
t(max)=0.2s

Δy=(u(y)t+0.5a(y)t^2)+0.2
Δy=(1.99*0.2-4.9*0.2^2)+0.2
Δy=0.2+0.2=0.4m

T2:
Δy=u(y)t+0.5a(y)t^2
0.4=0+4.9*t^2
t=0.29

THEREFORE TIME OF FLIGHT = 0.2+0.29=0.49s

Δx=u(x)*t
Δx=1.49*0.49
Δx=0.73m

v(y)=u(y)+a(y)*t (whole trajectory)
v(y)=1.99+9.8*0.49
v(y)=6.79m/s

*Using Pythagoras and trig which are assumed and elementary so I wont bother putting in the working*
Final Velocity= 6.95m/s 78° below the horizontal

thanks i understand everything except how u found the angle 78° below the horizontal

 

[Green Yoda]

thanks i understand everything except how u found the angle 78° below the horizontal

tan^-1(v(y)/u(y))
tan^-1(6.79/1.49)=78° (nearest degree)

 

[luo_ge]

tan^-1(v(y)/u(y))
tan^-1(6.79/1.49)=78° (nearest degree)

this might sound stupid but how do u know if something is below or above the horizontal?

 

[Green Yoda]

this might sound stupid but how do u know if something is below or above the horizontal?

When you draw the vector diagram..theta is below the horizontal line