Question

The solution to this question has been evading me for a while. This is technically a simple physics problem.

If a projectile is to be launched at a given angle theta, height 'h' and distance 'd'. Determine an equation to find the initial velocity.

Answer 1

By h and d, do you mean the starting height and distance or the maximum/final?
Typcially the way by which you solve this equation is to recognize that the velocity at the peak y-axis height is 0, then you can back solve. I may be overlooking something but that's how I would approach it if I knew more information

Answer 2

Probably looking for the equations, \[y=y_0+v_yt+\frac{1}{2}g t^2\]\[x=x_0+v_xt+\frac{1}{2}a t^2\]

Answer 3

A target is placed at 'd' metes away and 'h' meters high w.r.t the launch point. It must hit the target. An angle of launch is given. I need the velocity.

@espex
I don't know where the x0 and y0 variables come from. But, would vy = v sin theta and vx = v cos theta?

Also, in the x-direction there is no acceleration, so wouldn't it just be (vx)(t)?

Answer 4

okay i get what your question is now. give me a sec to type! :)

Answer 5

The zero (naught) point is your initial point that you are measuring from, vy is the velocity at point y. If you knew the initial velocity at angle theta then vsin(theta) would be the y component and vcos(theta) would be the x component.
As for the acceleration question, yes, in your example, you would just put a zero in there so that whole portion would drop out.

Answer 6

I see. But what would the final equation for v look like?

Answer 7

I believe you could use your angle and the x equation to solve for t and then solve the 'y' equation if you know your velocity.
\[y=VCos(θ)t+\frac{1}{2}g t^2 \space \space x=VSin(θ)t\]
I thought that amanda had the complete answer.

Answer 8

I believe I do? The idea behind that was to set up equations for both the x and y direction, solve for an expression with t, and then use that t expression to solve for V(o) using one of the questions. sorry i had to handwrite that hahaha the symbols and typing drive me insane . hope it helps!!!

Answer 9

Using the math program on here is a tad clunky. Good job on the solution. :)

Answer 10

:D

Answer 11

Thanks a a lot guys.
@Amandachen92 - for all that trouble!

Answer 12

Question though,

If I use d/v.cos, would it not give me the time it takes the projectile to reach the same level again?