Projectile Motion: Determining Velocity and Distance at Different Angles

[Charlotte57]

Homework Statement

find the initial velocity if a ball is shot at 0 degrees, goes 99.3 cm forward at a height of 25.5cm
then use this velocity to determine the x distance if shot at 60 degrees at a height of 114cm
part 1
Vinitial: ?
delta x: 99.3cm
delta y: 25.5cm
degree: 0
Part 2
Vinitial: from part 1
delta x: ?
delta y: 114cm
degree: 60
delta t: ?

Homework Equations

x = xinitial + Vinitial * t +.5at^2
x = -b+/- sq.rt.(b^2 - 4ac))/2a
others?

The Attempt at a Solution

99.3sqrt((2*25.5)/9.8 = -42.7cm/s
42.7 +/- sqrt 1823.29 - 4(-4.9)(114)/-9.8
-10.86 or 2.14
this is as far as I get bc I know t should be positive but 2.14 is way too large to be reasonable

 

[NFuller]

I'm not really sure what equations you are using in your work, particularly the first line. There are also issues with units, your value of g is in $\text{m/s}^{2}$ but your other values are in units of cm.

If you set up the one dimensional motion equation in the vertical direction, what do you get? Can you find the time of flight using this?

 

[Charlotte57]

Oh i didn't even see that the acceleration wasn't converted to cm! sorry for being confusing
and for the one dimensional motion equation do you mean this one?
V^2(final) = V^2(initial) + 2a(x(final) - x(initial)
but rearranged
if so I get:
-V^2(initial) = 2⋅980(99.3 - 0) - 0
V(initial) = -441.17cm/s
if I apply this to t using the quadratic formula
t = (441.17 +/- √(194628 - 4(490)(114))/2⋅490
t = 29.85s or -28.95s
yes?

 

[willem2]

You really should use more text. I think your first line finds the initial velocity in the right way. The answer is incorrect however, because you mix meters and cm.
You should really write something like this:
The equation for the height of the projectile is ... . Setting this to 0 and solving for t gives t = ..., Since [equation involving the x coordinatie of the projectile] the initial velocity is ...
I really have to idea what you try to do in the second line. You use the quadratic equation but I can't see why. What is the inital velociyy in the x and y direction? Give an equation for the x and y coordinate as a function of time.

 

[NFuller]

and for the one dimensional motion equation do you mean this one?

I mean the one involving $t$
$$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$$
What does this look like along the y (vertical) axis?