Needing some help about time of flight of a launched ball/horiz range

[AmandaP]

Homework Statement

Needing some help!!! Solution attempts are displayed

Relevant Equations

y0=uyt + 1/2gt^2

horizontal range: x=v * t

The parts in bold are the questions and also highlights of my attempts at the solutions for each number and letter portion of the question. I attached a word doc to show the diagram and it also has the same questions ;)

4) A launcher is setup at the edge of a table as shown below: (I attached the word document that shows the actual picture :) )

y is the vertical height of the launcher (above the floor), and vo is the launch speed of the ball. If y = 1.2 m, and vo = 5.1 m/s,

a) What is the time of flight of the ball?

b) What is the horizontal range of the ball?

See my solution attempts for question #4 below

Question #4
4a) y0=uyt +1/2 gt^2
1.2m = (0) t + (1/2)(9.80 m/s^2) t^2
1.2 = 0t +(4.9m/s^2)t^2
1.2m = 0 + 4.9m/s^2 t^2
4.9m/s^2 4.9m/s^2

I don't know how to draw a square root sign but I then took the square root of .24490 = t^2 in which the time of flight is t=0.494s

4b) x = v0 * t

v0=5.1 m/s
t = 0.494s

(5.1 m/s) * (0.494s)=2.51940 m

Horizontal range = 2.52 m (??)

Question #5
If the same launcher is angled upward 30° from the horizontal such that y and vo stay the same,

a) What is the time of flight of the ball?b) What is the horizontal range of the ball?

See my solution attempts for question #5 below
#5
a) 2v0 * sin2(theta)
g
2(5.1m/s) * sin(30) =0.52041 s (?)
(9.80m/s^2)

time of flight = 0.52041 s (unsure of this answer)

5b) x = v0 * t sin (theta)

5.1 m/s * (0.520s)(sin 30) = 1.3260 m (?)

 

[kuruman]

Your answers to question 4 look fine.

You should repeat the method you used for question 4 in question 5. The expression 2v0 * sin2(theta) gives the horizontal range only if the projectile returns to the same height from which it was launched. This is not the case here.

For posting equations, I recommend that you use LaTeX. Click on "LaTeX Guide" lower left corner to learn how. It's really easy.

 

[AmandaP]

That makes sense the feedback you gave me...I do feel confused about how I add the angle part then to my formulas??

If I use the same methods I did in part 4, then I will get the same answers, so I need to take into account the (sin 30) since that is the angle that the ball is being launched so would my new calculation be this?:

t=v0sin(30)
g
v0=5.2 m/s

(5.1 m/s) * sin(30) = 0.260 seconds (?)
9.80 m/s^2

So I removed the doubled part...I know I am overthinking this
Thank you for the LaTex feedback for inputting formulas

 

[PeroK]

That makes sense the feedback you gave me...I do feel confused about how I add the angle part then to my formulas??

If I use the same methods I did in part 4, then I will get the same answers, so I need to take into account the (sin 30) since that is the angle that the ball is being launched so would my new calculation be this?:

t=v0sin(30)
g
v0=5.2 m/s

(5.1 m/s) * sin(30) = 0.260 seconds (?)
9.80 m/s^2

I don't understand that calculation at all. Is $t$ the (total) time of flight?

 

[kuruman]

That makes sense the feedback you gave me...I do feel confused about how I add the angle part then to my formulas??

If I use the same methods I did in part 4, then I will get the same answers, so I need to take into account the (sin 30) since that is the angle that the ball is being launched so would my new calculation be this?:

t=v0sin(30)
g
v0=5.2 m/s

(5.1 m/s) * sin(30) = 0.260 seconds (?)
9.80 m/s^2

So I removed the doubled part...I know I am overthinking this
Thank you for the LaTex feedback for inputting formulas

No, you will not get the same answers. In question 4 the horizontal component of the velocity is 5.1 m/s and the vertical component is zero because the projectile is shot horizontally (I think). In question 5 the projectile is shot with the same speed, 5.1 m/s, but at angle 30° above the horizontal. Neither the horizontal nor the vertical components of the velocity are the same as in question 4. You need to write kinematic equations in both horizontal and vertical directions. Use the vertical equation to find the time of flight. Assume that you know the height of the cliff from question 4.

 

[AmandaP]

I don't understand that calculation at all. Is $t$ the (total) time of flight?

No, it is not, the format was shifted when it was posted. I scheduled an appointment with a tutor through my school. I am overthinking this right now and have to take a mental break :).

 

[haruspex]

If I use the same methods I did in part 4, then I will get the same answers

No, the method is the same, once you have understood what the first step in the method was, namely: Find the horizontal and vertical components of the initial velocity.
In 4, that was trivial; the initial vertical velocity was zero and the whole 5.1m/s was horizontal. In 5, you need to use trigonometry to find them.