Question

Molly kicks a soccer ball into the air with an initial velocity of 15 m/s. It lands 20 meters from where she kicked it. At what angle did Molly launch the ball?

Answer 1

`theta = 1/2 sin^-1(20/225) "radians"`

Explanation:

![https://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@7.61:20/Projectile-Motion]()

The x and y components of initial velocity `v_o = 15 m/s` are
1. `v_x = v_o cos theta;` and
2. `v_y = v_o sin theta - "gt"`
3. from 1) the distance in x is `x(t) = v_otcostheta`
a) Total distance in x, Range `R = 20= x(t_d) = v_ot_dcostheta`
b) Where `t_d` is the total distance required to travel R = 20 m
4. The displacement in y is `y(t) = v_o tsintheta - 1/2"gt"^2`
a) at time `t = t_d; y(t_d) = 0`
b) setting y=0 and solving for time, `t_d = 2v_osintheta/g`
5. Insert 4.a) into 3.a) we get, `R = 2v_o^2(costheta sintheta)/g`
a) 5. above can also be written as: `R = v_o^2/gsin2theta`

Now we know, `R = 20m; v_o = 15m/s` solve for `theta`
`theta = 1/2 sin^-1(20/225) "radians"`