Finding initial velocity and horizontal distance for a projectile with given vertical drop

[8320]

You are to find the initial velocity of the projectile, then use this velocity to calculate dx when dy=4.961m
given
dx=0.92m
dy=0.42
ay=9.81
v1y=0

2. Homework Equations
To find the time, you use the formula dy=v1y t + 1/2ay t2
You would also use this formula to find the initial velocity

3. The Attempt at a Solution
For the time, it would be the square root of 2dy over ay
t= 0.43s for the second part, t= 1.01s
then you put t= 0.43 into the above equation for the velocity...
this is where I am running into trouble...

what is the velocity?
it is this velocity that you use to find dx.
the equation in my textbook is dx=v1x t
I know that the dx will also be in the general area of 0.80-90m

 

[zgozvrm]

It seems you have some errors in your question.

I assume that v1y=0 is to mean that the initial velocity in the y direction ([itex]V_{iy}[/tex]) is 0 m/s.<br /> I also assume that by ay=9.81, you mean to say that the acceleration of gravity is assumed to be 9.81 [itex]m/s^2[/tex]. Since gravity is downward, then you should have [itex]a_y=-9.81 m/s^2[/tex]<br /> <br /> If that is true, then dy (the displacement in the y direction) can never be positive.Additionally, you have the formula for distance written incorrectly; it should be:<br /> $$d_y = V_{iy}t + \frac{1}{2}at^2 = V_{iy}t - 4.905t^2$$<br /> (note that there is no velocity value in the [itex]t^2[/tex] term.)[/itex][/itex][/itex][/itex]