A projectile is launched with an initial speed of 42 feet per second. It is projected at an angle of 50 degrees. How far does the projectile travel? How much farther does it travel if it is launched with an initial speed of 84 feet per second?

Question

whpalmer4 · Answer

can you find the vertical component of the initial velocity if the angle is 50 degrees (from the horizontal, presumably)?

anonymous · Answer

No, that is the exact words of the problem.

whpalmer4 · Answer

It never ceases to amaze me when people answer my questions with "those are the exact words of the problem" :-)  Here's a hint: you need to find the vertical component of the initial velocity :-)

whpalmer4 · Answer

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Can you find $v_x$ and $v_y$?

whpalmer4 · Answer

Once you find $v_y$, you're going to plug it into this formula:

$$h(t) = -\frac{1}{2}gt^2 + v_y t + h_0$$where $g=32\text{ ft/s}^2$ and
$h_0 = 0$ (because we are launching from the ground)

Solve that for the two values of $t$ that give $h(t) = 0$.  The difference between those values will be the time of flight.  With the time of flight in hand, you can compute the horizontal distance by using $x = v_x t$ where $t$ is the time of flight.

anonymous · Answer

Thank you, I will try and figure this out. My mind is just going blank..

whpalmer4 · Answer

tag me again if you just can't get through it.  I'm heading out for the night, but I'll check back in the morning.