Projectile Motion

[Vincezor]

Hey guys,

I've been having problems with two particular questions involving asymmetrical projectile motion.

1. A basketball aims to throw a basketball so it strikes a backboard 1m above the point of release with only a horizontal velocity. If the ball is shot with an initial speed of 4.8m/s, how far horizontally from the backboard must she stand?

EDIT: Worked it out!

Vertical Component:
to find t,

use to find u,

Now you know the total velocity is 4.8m/s, so you use pythag to find the horizontal initial velocity which = =



2. A stone is thrown with a speed of 18 m/s at an angle of 50 degrees above the horizontal from a bridge 15m above the surface of a wide river. Find the maximum height and range of the stone. At what speed does the stone splash into the water?

- I let u=18sin50, v=0, a=-10m/s/s, and used v^2 =U^2 + 2as to find answer for max height, then added 15 on top of that (24.51m)

I have no idea how to find the range though, or any of the subsequent questions.

[onur369]

Is Q1 approx 9.7m ?

[xZero]

1. The question said that the vertical height would be 1m then the vertical velocity is 0, use x = vt -0.5at^2 to find t then use t to find u. Now find the angle at which the basketball was shot at by usig the vertical velocity you just worked out and apply the equation x = ut

[onur369]

For Question 2 I got the max height as 24.5 if you add the 15 it is above the ground already, for its range i got 31.9.

If they are correct I can give full solution

[onur369]

For the range use the this formula if SYMMETRIC MOTION: Rmax= u^2 x sin2theta /10

[Vincezor]

For the range use the this formula if SYMMETRIC MOTION: Rmax= u^2 x sin2theta /10

Im pretty sure it isn't symmetrical motion, as the stone ends up 15m below from where it started. Also, the answer for the range is 42.2m (I think they used g=9.8 though)

[onur369]

Is this from a trial or something, I can possibly look it up

[Vincezor]

Is this from a trial or something, I can possibly look it up

Nope, just a booklet of questions we have to do for holiday homework...

[luffy]

Hey guys,

I've been having problems with two particular questions involving asymmetrical projectile motion.

1. A basketball aims to throw a basketball so it strikes a backboard 1m above the point of release with only a horizontal velocity. If the ball is shot with an initial speed of 4.8m/s, how far horizontally from the backboard must she stand?

- Don't know how to start this at all.

2. A stone is thrown with a speed of 18 m/s at an angle of 50 degrees above the horizontal from a bridge 15m above the surface of a wide river. Find the maximum height and range of the stone. At what speed does the stone splash into the water?

- I let u=18sin50, v=0, a=-10m/s/s, and used v^2 =U^2 + 2as to find answer for max height, then added 15 on top of that (24.51m)

I have no idea how to find the range though, or any of the subsequent questions.

Hello Vincezor

Question 1.
Vertically:
x = 1
v = 0
a = -10
u = ?

Find u (using v^2 = u^2 + 2ax) and you will find it is .

Find t, (using


 
Now draw a triangle to find horizontal initial velocity.
i.e. u_horizontal =


Therefore, horizontally:


a = 0

x = ?







Hope I did it right  


Question 2.

Your right, the max height should be 24.31m.

As for the range: (you cannot use range formula because they are at different heights)

Vertically: u = 18sin(50) m/s
              x = -15m
              a = -10




You would have to use quadratics to find the value of t (reject any negative t-values), which will be approximately be 3.59 seconds.

Now for the horizontal part of the motion:

u = 18cos(50)
v = 18cos(50)
a = 0
t 3.59
x = ? (i.e. range)





-> Is that the correct answer?

For the speed that it splashes into the water:
- v_horizontal = 18cos(50)

-v_vertical:







Therefore,



Is that right?

Hope I helped.

[Vincezor]

Hello Vincezor

Question 1.
Vertically:
x = 1
v = 0
a = -10
u = ?

Find u (using v^2 = u^2 + 2ax) and you will find it is .

Find t, (using


 
Now draw a triangle to find horizontal initial velocity.
i.e. u_horizontal =


Therefore, horizontally:


a = 0

x = ?







Hope I did it right  

I believe , which resulted in your answer being different from mine. (see above)

[luffy]

Hello Vincezor

Question 1.
Vertically:
x = 1
v = 0
a = -10
u = ?

Find u (using v^2 = u^2 + 2ax) and you will find it is .

Find t, (using


 
Now draw a triangle to find horizontal initial velocity.
i.e. u_horizontal =


Therefore, horizontally:


a = 0

x = ?







Hope I did it right 

I believe , which resulted in your answer being different from mine. (see above)

Sorry, my mistake. I'll edit my original post.

[schnappy]

2.
Find the time of the flight from the vertical: u=18sin(50), x=-15, a=-10
Then horizontally:
v=d/t, v=18cos(50), t=[as above], d=?
Where d is the range of your motion.

For the final speed, the horizontal component is just 18cos(50). Use the kinematics equations to find the final velocity when x=-15, then create a right angled triangle.