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I have a problem where I am trying to find the displacement in the x axis (i.e distance along the ground) of a projectile from the start point (where it is launched) to the point where it reaches a given height.

I know the initial velocity (v0), the launch angle, the target height (h) and gravitational acceleration (G). Air resistance can be ignored.

I have been able to find equations for calculating this given the time it takes for the projectile to reach the height (h), however I do not know the time taken.

Thanks in advance for any help.

1 Answers 1

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The vertical component of the initial velocity is $v_0 \sin \alpha$, while the horizontal component is $v_0 \cos \alpha$. Vertically, the motion is accelerated with acceleration $-g$, so the height at time $t$ is

$$h(t) = v_0 (\sin \alpha) t - \frac{1}{2} gt^2.$$

Now fix your height $h_0$ and find $t_0$ such that

$$h(t_0) = h_0.$$

$t_0$ is the time taken to reach the height $h_0$.

Horizontally, the motion is uniform, and the horizontal distance at time $t$ is given simply by

$$x(t) = v_0 (\cos \alpha)t$$

Therefore the horizontal distance that the body covers is $x(t_0)$.

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    Thanks so much Stefano, this was really helpful. – 2017-03-02