- #1

AmandaP

- 3

- 1

- 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 ;)

1.2m = (0) t + (1/2)(9.80 m/s^2) t^2

1.2 = 0t +(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

v0=5.1 m/s

t = 0.494s

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

Horizontal range = 2.52 m (??)

g

(9.80m/s^2)

time of flight = 0.52041 s

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

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

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

b) What is the horizontal range of the ball?

Question #4

4a)y0=uyt +1/2 gt^2*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)

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^24.9m/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 * tv0=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

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

#5

a)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 (?)