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basic kinematics M2 question, where did i go wrong?

The object of a game is to throw a ball B from a point A to hit a target T which is placed at the top of a vertical pole, as shown in Figure 3. The point A is 1 m above horizontal ground and the height of the pole is 2 m. The pole is at a horizontal distance of 10 m from A. The ball B is projected from A with a speed of 11 m s–1 at an angle of elevation of 30 degrees. The ball hits the pole at the point C. The ball B and the target T are modelled as particles.

my working:

U_x = 11cos30
a_x = 0
S_x = 5
V_x = 0
t_x = t

5 = (0.5)[(11cos30)/t]
therefore t = 11cos30/10
t = 0.9526279442
2t = 1.905 (3dp). this is the time taken for the whole journey

U_y = 11sin30
a_y = 9.8
S_y = 1
t_y = t

s = ut - (at^2)/2
4.9t^2 - t(11sin30) + 1 = 0
therefore t = 4.3817 and 0.2281

which one of these t values (for the y components) do i use?

any help would be much appreciated :smile:
Reply 1
Avatar for sulexk
sulexk
11
Original post by Lololo
The object of a game is to throw a ball B from a point A to hit a target T which is placed at the top of a vertical pole, as shown in Figure 3. The point A is 1 m above horizontal ground and the height of the pole is 2 m. The pole is at a horizontal distance of 10 m from A. The ball B is projected from A with a speed of 11 m s–1 at an angle of elevation of 30 degrees. The ball hits the pole at the point C. The ball B and the target T are modelled as particles.

my working:

U_x = 11cos30
a_x = 0
S_x = 5
V_x = 0
t_x = t

5 = (0.5)[(11cos30)/t]
therefore t = 11cos30/10
t = 0.9526279442
2t = 1.905 (3dp). this is the time taken for the whole journey

U_y = 11sin30
a_y = 9.8
S_y = 1
t_y = t

s = ut - (at^2)/2
4.9t^2 - t(11sin30) + 1 = 0
therefore t = 4.3817 and 0.2281

which one of these t values (for the y components) do i use?

any help would be much appreciated :smile:


Hello, sorry to ask, but is the question asking how far up does the ball hit the pole?

Thank you
Reply 2
Avatar for Lololo
Lololo
OP
Original post by sulexk
Hello, sorry to ask, but is the question asking how far up does the ball hit the pole?

Thank you
no, it tells you that the pole is 2 meters long. it's asking for the time it takes to get from A to T (the top of the pole)
Reply 3
Avatar for sulexk
sulexk
11
Oh I see, I shall work it out again
Original post by Lololo
The object of a game is to throw a ball B from a point A to hit a target T which is placed at the top of a vertical pole, as shown in Figure 3. The point A is 1 m above horizontal ground and the height of the pole is 2 m. The pole is at a horizontal distance of 10 m from A. The ball B is projected from A with a speed of 11 m s–1 at an angle of elevation of 30 degrees. The ball hits the pole at the point C. The ball B and the target T are modelled as particles.

my working:

U_x = 11cos30
a_x = 0
S_x = 5
V_x = 0
t_x = t

5 = (0.5)[(11cos30)/t]
therefore t = 11cos30/10
t = 0.9526279442
2t = 1.905 (3dp). this is the time taken for the whole journey

U_y = 11sin30
a_y = 9.8
S_y = 1
t_y = t

s = ut - (at^2)/2
4.9t^2 - t(11sin30) + 1 = 0
therefore t = 4.3817 and 0.2281

which one of these t values (for the y components) do i use?

any help would be much appreciated :smile:


I've seen this question before, but I don't think I have it with me.

It says that the ball hits the pole at C (not at the target T).

Since we believe that it ends up at a different height from when it started, Vx will not equal 0 when Sx=5.

Use Sx =10 to get a value for t. Then use that in the vertical equation to find the height of C above the starting place.
Reply 5
Avatar for sulexk
sulexk
11
Original post by Lololo
The object of a game is to throw a ball B from a point A to hit a target T which is placed at the top of a vertical pole, as shown in Figure 3. The point A is 1 m above horizontal ground and the height of the pole is 2 m. The pole is at a horizontal distance of 10 m from A. The ball B is projected from A with a speed of 11 m s–1 at an angle of elevation of 30 degrees. The ball hits the pole at the point C. The ball B and the target T are modelled as particles.

my working:

U_x = 11cos30
a_x = 0
S_x = 5
V_x = 0
t_x = t

5 = (0.5)[(11cos30)/t]
therefore t = 11cos30/10
t = 0.9526279442
2t = 1.905 (3dp). this is the time taken for the whole journey

U_y = 11sin30
a_y = 9.8
S_y = 1
t_y = t

s = ut - (at^2)/2
4.9t^2 - t(11sin30) + 1 = 0
therefore t = 4.3817 and 0.2281

which one of these t values (for the y components) do i use?

any help would be much appreciated :smile:



Hey Lololo,

I got an answer of 1.05 s

It is known that the horizontal distance between the pole and the point of release(A) is 10m.

so we use
s=10
u= 11cos30
v
a=0
t=?

we know the horizontal velocity to be 11cos30(ms^-1) as it is released from an angle of elevation of 30 degrees, and the horizontal component of velocity is 11cos30.

It does not accelerate in the horizontal direction, this is the case with a projectile,
so a=0

then use the equation s=ut + 1/2at^2
as a=0, this breaks down to: s=ut

s=ut

now s=10
u=11cos30

and we need to find t. then t=10/11cos30 = 1.05

I hope this helps!

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