Yes, I did this.Brianna I said:Ok, when I was initially taught this I was told to make a chart in both the x and y with what you know and do not know.
The setup would be something like this:
______ X _____ Y
a =
vi =
d =
t =
From there try to solve for maybe t using the y-variables (since you don't have that either) and then plugging in t in an equation that will give you Vix.
I have not gotten any of it done because I have no idea which formula to use. All I know is that on my horizontal I have d = 200 m and a = 0. Then for my vertical I have d = 0 and a = 9.8m/s^2(down). I don't have time and all the formulas I was given deal with time which is why I am confusedgneill said:You need to show what you've done before we can help.
Veronica_Oles said:I have not gotten any of it done because I have no idea which formula to use. All I know is that on my horizontal I have d = 200 m and a = 0. Then for my vertical I have d = 0 and a = 9.8m/s^2(down). I don't have time and all the formulas I was given deal with time which is why I am confused
Yes I know those.haruspex said:For motion under constant acceleration, there are five standard variables (usually known as SUVAT, s for displacement, u for initial speed, v for final speed, a for acceleration, t for time). There are correspondingly five equations. Each equation omits one of the five variables. Are you familiar with these?
Since you do not care about time, pick one that does not involve time.Veronica_Oles said:Yes I know those.
haruspex said:Since you do not care about time, pick one that does not involve time.
The only one I see that doesn't involve time is vf^2 = vi^2 + 2adharuspex said:Since you do not care about time, pick one that does not involve time.
Sorry, I wasn't thinking clearly. You do care about time, but you have to start with the horizontal direction. If it is hit with speed v, how long to reach ythat distance?Veronica_Oles said:The only one I see that doesn't involve time is vf^2 = vi^2 + 2ad
T= 200/ViCos 45 ??!haruspex said:Sorry, I wasn't thinking clearly. You do care about time, but you have to start with the horizontal direction. If it is hit with speed v, how long to reach ythat distance?
Yes.Veronica_Oles said:T= 200/ViCos 45 ??!
What is the initial vertical speed? How long before it lands?Simon Dobbs said:still confusing me
if u is the initial velocity at an angle θ with the horizontalVeronica_Oles said:A baseball player hits a ball 200.0m home run. The ball travels at angle of 45° with the horizontal just after being hit. Determine the initial speed. Assume the ball lands the same height it was hit.
A projectile launcher is a device used to launch objects in a particular direction and at a specific angle. It is commonly used for experiments in physics to study the motion of objects.
A projectile launcher works by using a spring or compressed air to create a force that propels an object forward. The angle of the launcher can be adjusted to change the trajectory of the object.
The optimal angle for launching a baseball at a 45 degree angle to the horizontal would be 45 degrees. This angle allows for the maximum distance and height to be achieved, as well as a balanced trajectory.
Air resistance, also known as drag, can affect the trajectory of a baseball launched at a 45 degree angle. As the ball travels through the air, it experiences a force in the opposite direction of its motion, causing it to slow down and potentially change its trajectory.
Other factors that can affect the trajectory of a baseball launched at a 45 degree angle include the initial velocity, the mass and size of the ball, and external forces such as wind or air resistance. The surface on which the ball lands can also impact its trajectory.