Projectile Motion Question (1 Viewer)

luo_ge

Member
Joined
Sep 26, 2016
Messages
60
Gender
Male
HSC
2016
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

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
Need more info on the actual motion of the projectile...where its thrown from and in what direction
 

luo_ge

Member
Joined
Sep 26, 2016
Messages
60
Gender
Male
HSC
2016
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

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
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

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
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

Hi Φ
Joined
Mar 28, 2015
Messages
2,859
Gender
Male
HSC
2017
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

Member
Joined
Sep 26, 2016
Messages
60
Gender
Male
HSC
2016
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
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top