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?

1 Answer

Answer:

#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"#