projectile motion help (1 Viewer)

[karen88]

Projectile Motion question i dun get....

At the highest point of its path a body has a velocity of 10m/s and is 8m above the ground. Find the angle of projection and the initial velocity.

ive found the initial velocity which is 16m/s but i can't find the angle of projection... what did u guys get... i need help i keep on getting 35 degrees 15 min... and the answer says 51 degrees 23 mins...

 

[Templar]

acceleration in y axis = -g
velocity in y axis = -gt
y=-1/2gt^2+8

Solve for y=0, t=4/sqrt(g)

Velocity in y axis at impact = 4sqrt(g)
Velocity in x axis at impact = 10

tan@=4sqrt(g)/10

@=51 degrees

Since launch and impact angles are the same for constant acceleration, launch angle=51

 

[karen88]

how did u get the velocity in y axis.... i dun noe how!!!

 

[Templar]

Acceleration in y axis is -g. Integrate. Original velocity in y axis is 0 (object is at max height), so C=0.

 

[karen88]

noo... i get that... it was the other one... the one on impact...

 

[king_of_boredom]

noo... i get that... it was the other one... the one on impact...

so are you asking for angle of projection or angle of impact?

 

[acmilan]

To get the y-component of velocity on impact you insert t into the formula for the y-velocity, where t is the total time of the flight

 

[karen88]

angle of projection.. it said in the question... i was just wondering how templar got 4sqrt(g) for the velocity of y axis on impact!

i did that!!! it didn't get the same answer .....
could someone show me the whole working out... im really bad with projectile....

 

[Templar]

so are you asking for angle of projection or angle of impact?

In this case they're the same.

y is given as a formula of t. Solve for y=0 (impact), you should get some value of t (4/sqrt(g)).

Sub t into formula for vertical velocity. Vertical velocity = -4*sqrt(g)

 

[karen88]

omg... i feel so stupid... im going to fail projectile motion... i still dun get it....

 

[Templar]

accel=-g
v=int (-g) dt
=-gt+C

At t=0, v=0. Therefore C=0

v=-gt
y=int (-gt) dt
=-1/2gt^2+k

At t=0, y=8. Therefore k=8

At impact, y=0. So solve for t when y=0

1/2gt^2=8
gt^2=16
t^2=16/g
t=4/sqrt(g)

To find v at impact, sub t=4/sqrt(g) into formula for v.

v=-g(4/sqrt(g))
=-sqrt(g)*4

Then use trig (tan) to get the angle, @=arctan(-y/x)

 

[karen88]

o9hhhhhh!!!!! i get it now!!!!! thnx so much!!!!!i feel so stupid.... the projectile motion is killing me.... i get all of my other topics and aces them.... except i have no clue with projectile motion, prob becos theres so many formulas and so many ways u could do them.... really pissing me off...