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?

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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?

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Answer 1 · 2016-02-07T19:36:22

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

Explanation:

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

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

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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

Explanation:

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