I started learning Kinematics for fun at Khan Academy, and so far it's quite interesting. Anyway, I was reading about 2D Projectile Motion, and was trying to solve the second example myself. You can find it by searching Pumpkin launched at an angle.

An air cannon is used to launch a pumpkin off a cliff of height $H=18.0m$ with an initial speed $v_0=11.4 \frac{m}{s}$ at an angle of $\theta=52.1$ What is the speed of the pumpkin right before it hits the ground?

I tried applying the formula

$$\Delta y = v_{y0}t + \frac{1}{2}at^2$$

Now, I have the initial vertical speed, we know that the vertical acceleration is constant, so I've found $t=3.04$ and then plugged it into $$\Delta s = (\frac{v_0+v_f}{2})t$$ and found that $v_f = 3\frac{m}{s}$

Turns it it's completely wrong. Would love to know where my mistake was.

Another thing that bothered me: when I was calculating the time I ended up with a 2nd degree equation, which meant that one of the solutions for $t$ was negative, already suspicious. Is this a part of the mistake? Can that be, and if so, does it have a deeper meaning?


closed as off-topic by ZeroTheHero, Kyle Kanos, Buzz, Aaron Stevens, John Rennie Feb 6 at 7:57

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Hint: Try breaking the launch velocity into its components. If initial velocity vector is $\overline {v_0}$ inclined at $\theta$ to the horizontal, it's horizontal and vertical scalar components will be $v_0\cos \theta$ and $v_0 \sin \theta$ respectively. Then make use of the equations of projectile motion. You've calculated time correctly.

Now to find the final vertical velocity, make use of $v_y=u\sin\theta-gt$. Now note that you have the final velocities in the horizontal and vertical directions. Simply calculate the magnitude of the resultant final velocity vector as $\sqrt{v_x^2+v_y^2}$ and you're done.

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