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A 11.54 kg cantaloupe takes 0.651s to launch, is launched off a 22.87 m-cliff with a resultant Force of 236.55 N. at an angle of at an angle 57.66° above the horizontal. Given gravity and Air resistance are accounted for at launch. What is the launch velocity of the Y axis?

User James Brightman
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1 Answer

23 votes
23 votes

We know that

• The mass is 11.54 kg.

,

• The time taken is 0.651 seconds.

,

• The distance covered is 22.87 meters.

,

• The resutant force is 236.55 N.

,

• The angle of launch is 57.66.

Let's use Newton's Second Law.


\Sigma F_y=ma\to F-W=ma

Where F = 236.55N, W = mg, m = 11.54kg, and g = 9.81 m/s^2.


\begin{gathered} 236.55N-11.54\operatorname{kg}\cdot9.81\cdot(m)/(s^2)=11.54\operatorname{kg}\cdot a \\ 236.55N-113.21N=11.54\operatorname{kg}\cdot a \\ a=\frac{123.34N}{11.54\operatorname{kg}} \\ a\approx10.7\cdot(m)/(s^2) \end{gathered}

The acceleration of the object is 10.7 m/s^2.

Now, let's find the initial vertical velocity.


\begin{gathered} v_y=v_(oy)+gt \\ 0=v_(oy)-9.8\cdot0.651\sec \\ v_(0y)=6.4\cdot(m)/(s) \end{gathered}

The initial vertical velocity is 6.4 m/s.

User Shanikqua
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3.1k points