# how to find the horizontal position where a projectile lands with inital height, initial velocity and angle

I am trying to find the horizontal position where a projectile lands. I have been given the initial height, initial velocity and the angle in degrees the projectile is launched at.

I have tried finding the vertical and horizontal velocity and then finding time using v = u + at and then finding the distance from v x t but this didn't give me the right answer.

This is what I have tried

• You probably mean the horizontal distance traveled, not the distance along the arc of travel, right? – MPW Jul 2 '18 at 17:12
• the horizontal velocity is going to be constant, you need to find time. Post your full solution and we'll take a look – Vasya Jul 2 '18 at 17:28

## 2 Answers

You must decompose your problem into a horizontal motion with no acceleration, and a vertical moment with only gravity. The total time of flight is given by the vertical motion. You launch the projectile with a certain vertical velocity (the vertical component of your original velocity). It will go up, then down. When it reaches ground, the motion stops. Calculate this time. On the horizontal axis, the projectile moves the same amount of time as in vertical direction, but there is no acceleration. The constant speed is the horizontal component of the initial velocity.

Your $t_1$ calculation is just for the upward motion to reach max height. You need to calculate the time $t_2$ it takes to go from max height to the ground which is further than the upward motion. Max height will be:

$h_{max} = h_0 + \frac{v_y}{2}\cdot t_1$

$t_2 = \sqrt{\frac{2\cdot h_{max}}{a}}$

Distance $x = v_x\cdot (t_1 + t_2)$