# Thread: calculating initial velocity(only given angle and distance traveled)

1. ## calculating initial velocity(only given angle and distance traveled)

ok here is the problem:
A projectile is shot up at an angle of 35degrees. it strikes the ground at a horizontal distance of 4km. calculate the initial velocity.
I'm not sure where to start on this one given the above information, can someone please point me in the right direction? thanks.

2. ## Physics

You need to know some basic physics, namely the kinematic equations. The only force is gravity, acting downwards (negative y direction), producing an acceleration g=9.8 m/s^2. Thus the position as a function of time, using our start as the origin, is given by $x(t)=v_{0x}t$, $y(t)=v_{0y}t-\frac{1}{2}gt^2$, where $v_{0x}$ and $v_{0y}$ are the x and y components of the launch velocity, respectively.
y(t)=0 when the projectile hits the ground, so solve that for t (there is a t=0 solution, which is because the projectile is launched from the ground; you want the nonzero solution). Plug this into x(t) to get the range as a function of $v_{0x}$, $v_{0y}$, and g. $v_{0x}$ and $v_{0y}$ can in turn be expressed in terms of the launch velocity $v_0$ and launch angle. You thus have an equation relating the unknown $v_0$ to the known range, launch angle, and acceleration of gravity.
--Kevin C.

3. i think i might have the answer:
i used 4000m = (vinitial^2 *sin2theta)/9.8
and solved for vinitial and get 204.24438m/s
can someone verify that this is correct, thanks.

4. Originally Posted by ihaveamathproblem
i think i might have the answer:
i used 4000m = (vinitial^2 *sin2theta)/9.8
and solved for vinitial and get 204.24438m/s
can someone verify that this is correct, thanks.
That is correct. Good job!

-Dan

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# find velocity given angle and distance

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